{"id":409,"date":"2021-02-04T02:02:24","date_gmt":"2021-02-04T02:02:24","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=409"},"modified":"2022-03-16T05:48:22","modified_gmt":"2022-03-16T05:48:22","slug":"drawing-graphs-of-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/drawing-graphs-of-functions\/","title":{"raw":"Drawing Graphs of Functions","rendered":"Drawing Graphs of Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Analyze a function and its derivatives to draw its graph<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h3>Guidelines for Graphing a Function<\/h3>\r\n<p id=\"fs-id1165042602938\">We now have enough analytical tools to draw graphs of a wide variety of algebraic and transcendental functions. Before showing how to graph specific functions, let\u2019s look at a general strategy to use when graphing any function.<\/p>\r\n\r\n<div id=\"fs-id1165042602944\" class=\"textbox examples\">\r\n<h3>Problem-Solving Strategy: Drawing the Graph of a Function<\/h3>\r\n<p id=\"fs-id1165042602951\">Given a function [latex]f[\/latex] use the following steps to sketch a graph of [latex]f[\/latex]:<\/p>\r\n\r\n<ol id=\"fs-id1165042602969\">\r\n \t<li>Determine the domain of the function.<\/li>\r\n \t<li>Locate the [latex]x[\/latex]- and [latex]y[\/latex]-intercepts.<\/li>\r\n \t<li>Evaluate [latex]\\underset{x\\to \\infty }{\\lim}f(x)[\/latex] and [latex]\\underset{x\\to \u2212\\infty }{\\lim}f(x)[\/latex] to determine the end behavior. If either of these limits is a finite number [latex]L[\/latex], then [latex]y=L[\/latex] is a horizontal asymptote. If either of these limits is [latex]\\infty [\/latex] or [latex]\u2212\\infty[\/latex], determine whether [latex]f[\/latex] has an oblique asymptote. If [latex]f[\/latex] is a rational function such that [latex]f(x)=\\frac{p(x)}{q(x)}[\/latex], where the degree of the numerator is greater than the degree of the denominator, then [latex]f[\/latex] can be written as\r\n<div id=\"fs-id1165042617521\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=\\dfrac{p(x)}{q(x)}=g(x)+\\dfrac{r(x)}{q(x)}[\/latex],<\/div>\r\nwhere the degree of [latex]r(x)[\/latex] is less than the degree of [latex]q(x)[\/latex]. The values of [latex]f(x)[\/latex] approach the values of [latex]g(x)[\/latex] as [latex]x\\to \\pm \\infty[\/latex]. If [latex]g(x)[\/latex] is a linear function, it is known as an <em>oblique asymptote<\/em>.<\/li>\r\n \t<li>Determine whether [latex]f[\/latex] has any vertical asymptotes.<\/li>\r\n \t<li>Calculate [latex]f^{\\prime}[\/latex]. Find all critical points and determine the intervals where [latex]f[\/latex] is increasing and where [latex]f[\/latex] is decreasing. Determine whether [latex]f[\/latex] has any local extrema.<\/li>\r\n \t<li>Calculate [latex]f^{\\prime \\prime}[\/latex]. Determine the intervals where [latex]f[\/latex] is concave up and where [latex]f[\/latex] is concave down. Use this information to determine whether [latex]f[\/latex] has any inflection points. The second derivative can also be used as an alternate means to determine or verify that [latex]f[\/latex] has a local extremum at a critical point.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<p id=\"fs-id1165042617762\">Now let\u2019s use this strategy to graph several different functions. We start by graphing a polynomial function.<\/p>\r\n\r\n<div id=\"fs-id1165042617768\" class=\"textbook exercises\">\r\n<h3>Example: Sketching a Graph of a Polynomial<\/h3>\r\nSketch a graph of [latex]f(x)=(x-1)^2 (x+2)[\/latex]\r\n<div id=\"fs-id1165042617770\" class=\"exercise\">[reveal-answer q=\"fs-id1165042707761\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042707761\"]\r\n<p id=\"fs-id1165042707761\">Step 1. Since [latex]f[\/latex] is a polynomial, the domain is the set of all real numbers.<\/p>\r\n<p id=\"fs-id1165042707768\">Step 2. When [latex]x=0, \\, f(x)=2[\/latex]. Therefore, the [latex]y[\/latex]-intercept is [latex](0,2)[\/latex]. To find the [latex]x[\/latex]-intercepts, we need to solve the equation [latex](x-1)^2 (x+2)=0[\/latex], which gives us the [latex]x[\/latex]-intercepts [latex](1,0)[\/latex] and [latex](-2,0)[\/latex]<\/p>\r\n<p id=\"fs-id1165042707901\">Step 3. We need to evaluate the end behavior of [latex]f[\/latex]. As [latex]x\\to \\infty[\/latex], [latex](x-1)^2 \\to \\infty [\/latex] and [latex](x+2)\\to \\infty[\/latex]. Therefore, [latex]\\underset{x\\to \\infty }{\\lim}f(x)=\\infty[\/latex]. As [latex]x\\to \u2212\\infty[\/latex], [latex](x-1)^2 \\to \\infty [\/latex] and [latex](x+2) \\to \u2212\\infty[\/latex]. Therefore, [latex]\\underset{x\\to -\\infty }{\\lim}f(x)=\u2212\\infty[\/latex]. To get even more information about the end behavior of [latex]f[\/latex], we can multiply the factors of [latex]f[\/latex]. When doing so, we see that<\/p>\r\n\r\n<div id=\"fs-id1165042525463\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=(x-1)^2 (x+2)=x^3-3x+2[\/latex]<\/div>\r\n<p id=\"fs-id1165042525532\">Since the leading term of [latex]f[\/latex] is [latex]x^3[\/latex], we conclude that [latex]f[\/latex] behaves like [latex]y=x^3[\/latex] as [latex]x\\to \\pm \\infty[\/latex].<\/p>\r\n<p id=\"fs-id1165043382921\">Step 4. Since [latex]f[\/latex] is a polynomial function, it does not have any vertical asymptotes.<\/p>\r\n<p id=\"fs-id1165043382928\">Step 5. The first derivative of [latex]f[\/latex] is<\/p>\r\n\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=3x^2-3[\/latex]<\/div>\r\n<p id=\"fs-id1165043382970\">Therefore, [latex]f[\/latex] has two critical points: [latex]x=1,-1[\/latex]. Divide the interval [latex](\u2212\\infty ,\\infty)[\/latex] into the three smaller intervals: [latex](\u2212\\infty ,-1)[\/latex], [latex](-1,1)[\/latex], and [latex](1,\\infty )[\/latex]. Then, choose test points [latex]x=-2[\/latex], [latex]x=0[\/latex], and [latex]x=2[\/latex] from these intervals and evaluate the sign of [latex]f^{\\prime}(x)[\/latex] at each of these test points, as shown in the following table.<\/p>\r\n\r\n<table id=\"fs-id1165043383127\" class=\"unnumbered\" summary=\"This table has four rows and four columns. The first row is a header row, and it reads Interval, Test Point, Sign of Derivative f\u2019(x) = 3x2 \u2013 3 = 3(x \u2013 1)(x + 1), and Conclusion. Under the header row, the first column reads (\u2212\u221e, \u22121), (\u22121, 1), and (1, \u221e). The second column reads x = \u22122, x = 0, and x = 2. The third column reads (+)(\u2212)(\u2212) = +, (+)(\u2212)(+) = \u2212, and (+)(+)(+) = +. The fourth column reads f is increasing, f is decreasing, and f is increasing.\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Interval<\/th>\r\n<th>Test Point<\/th>\r\n<th>Sign of Derivative [latex]f^{\\prime}(x)=3x^2-3=3(x-1)(x+1)[\/latex]<\/th>\r\n<th>Conclusion<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex](\u2212\\infty ,-1)[\/latex]<\/td>\r\n<td>[latex]x=-2[\/latex]<\/td>\r\n<td>[latex](+)(\u2212)(\u2212)=+[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is increasing.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](-1,1)[\/latex]<\/td>\r\n<td>[latex]x=0[\/latex]<\/td>\r\n<td>[latex](+)(\u2212)(+)=\u2212[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is decreasing.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](1,\\infty )[\/latex]<\/td>\r\n<td>[latex]x=2[\/latex]<\/td>\r\n<td>[latex](+)(+)(+)=+[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is increasing.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165042539204\">From the table, we see that [latex]f[\/latex] has a local maximum at [latex]x=-1[\/latex] and a local minimum at [latex]x=1[\/latex]. Evaluating [latex]f(x)[\/latex] at those two points, we find that the local maximum value is [latex]f(-1)=4[\/latex] and the local minimum value is [latex]f(1)=0[\/latex].<\/p>\r\n<p id=\"fs-id1165042539284\">Step 6. The second derivative of [latex]f[\/latex] is<\/p>\r\n\r\n<div id=\"fs-id1165042539292\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime \\prime}(x)=6x[\/latex]<\/div>\r\n<p id=\"fs-id1165042539318\">The second derivative is zero at [latex]x=0[\/latex]. Therefore, to determine the concavity of [latex]f[\/latex], divide the interval [latex](\u2212\\infty ,\\infty)[\/latex] into the smaller intervals [latex](\u2212\\infty ,0)[\/latex] and [latex](0,\\infty )[\/latex], and choose test points [latex]x=-1[\/latex] and [latex]x=1[\/latex] to determine the concavity of [latex]f[\/latex] on each of these smaller intervals as shown in the following table.<\/p>\r\n\r\n<table id=\"fs-id1165042381927\" class=\"unnumbered\" summary=\"This table has three rows and four columns. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019\u2019(x) = 6x, and Conclusion. Under the header row, the first column reads (\u2212\u221e, 0) and (0, \u221e). The second column reads x = \u22121 and x = 1. The third column reads \u2212 and +. The fourth column reads f is concave down and f is concave up.\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Interval<\/th>\r\n<th>Test Point<\/th>\r\n<th>Sign of [latex]f^{\\prime \\prime}(x)=6x[\/latex]<\/th>\r\n<th>Conclusion<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex](\u2212\\infty ,0)[\/latex]<\/td>\r\n<td>[latex]x=-1[\/latex]<\/td>\r\n<td>[latex]-[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is concave down.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](0,\\infty )[\/latex]<\/td>\r\n<td>[latex]x=1[\/latex]<\/td>\r\n<td>[latex]+[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is concave up.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165042382153\">We note that the information in the preceding table confirms the fact, found in step 5, that [latex]f[\/latex] has a local maximum at [latex]x=-1[\/latex] and a local minimum at [latex]x=1[\/latex]. In addition, the information found in step 5\u2014namely, [latex]f[\/latex] has a local maximum at [latex]x=-1[\/latex] and a local minimum at [latex]x=1[\/latex], and [latex]f^{\\prime}(x)=0[\/latex] at those points\u2014combined with the fact that [latex]f^{\\prime \\prime}[\/latex] changes sign only at [latex]x=0[\/latex] confirms the results found in step 6 on the concavity of [latex]f[\/latex].<\/p>\r\n<p id=\"fs-id1165042709981\">Combining this information, we arrive at the graph of [latex]f(x)=(x-1)^2 (x+2)[\/latex] shown in the following graph.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"400\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211132\/CNX_Calc_Figure_04_06_015.jpg\" alt=\"The function f(x) = (x \u22121)2 (x + 2) is graphed. It crosses the x axis at x = \u22122 and touches the x axis at x = 1.\" width=\"400\" height=\"347\" \/> Figure 23.[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Sketching a Graph of a Polynomial.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/QhOY9VfIefo?controls=0&amp;start=003&amp;end=317&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.6LimitsAtInfinityAndAsymptotesPart2_3to317_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.6 Limits at Infinity and Asymptotes (part 2 - curve sketching)\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1165042710046\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSketch a graph of [latex]f(x)=(x-1)^3 (x+2)[\/latex]\r\n<div id=\"fs-id1165042710050\" class=\"exercise\">\r\n\r\n[reveal-answer q=\"80046723\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"80046723\"]\r\n<p id=\"fs-id1165042710130\">[latex]f[\/latex] is a fourth-degree polynomial.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165042710106\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042710106\"]\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"401\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211136\/CNX_Calc_Figure_04_06_028.jpg\" alt=\"The function f(x) = (x \u22121)3(x + 2) is graphed.\" width=\"401\" height=\"520\" \/> Figure 24.[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042710140\" class=\"textbook exercises\">\r\n<h3>Example: Sketching a Rational Function<\/h3>\r\nSketch the graph of [latex]f(x)=\\dfrac{x^2}{1-x^2}[\/latex]\r\n<div id=\"fs-id1165042710142\" class=\"exercise\">[reveal-answer q=\"fs-id1165042407414\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042407414\"]\r\n<p id=\"fs-id1165042407414\">Step 1. The function [latex]f[\/latex] is defined as long as the denominator is not zero. Therefore, the domain is the set of all real numbers [latex]x[\/latex] except [latex]x=\\pm 1[\/latex].<\/p>\r\n<p id=\"fs-id1165042407440\">Step 2. Find the intercepts. If [latex]x=0[\/latex], then [latex]f(x)=0[\/latex], so 0 is an intercept. If [latex]y=0[\/latex], then [latex]\\frac{x^2}{1-x^2}=0[\/latex], which implies [latex]x=0[\/latex]. Therefore, [latex](0,0)[\/latex] is the only intercept.<\/p>\r\n<p id=\"fs-id1165042407553\">Step 3. Evaluate the limits at infinity. Since [latex]f[\/latex] is a rational function, divide the numerator and denominator by the highest power in the denominator: [latex]x^2[\/latex]. We obtain<\/p>\r\n\r\n<div id=\"fs-id1165042407571\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\pm \\infty }{\\lim}\\frac{x^2}{1-x^2}=\\underset{x\\to \\pm \\infty }{\\lim}\\frac{1}{\\frac{1}{x^2}-1}=-1[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042407655\">Therefore, [latex]f[\/latex] has a horizontal asymptote of [latex]y=-1[\/latex] as [latex]x\\to \\infty [\/latex] and [latex]x\\to \u2212\\infty[\/latex].<\/p>\r\n<p id=\"fs-id1165042491004\">Step 4. To determine whether [latex]f[\/latex] has any vertical asymptotes, first check to see whether the denominator has any zeroes. We find the denominator is zero when [latex]x=\\pm 1[\/latex]. To determine whether the lines [latex]x=1[\/latex] or [latex]x=-1[\/latex] are vertical asymptotes of [latex]f[\/latex], evaluate [latex]\\underset{x\\to 1}{\\lim}f(x)[\/latex] and [latex]\\underset{x\\to \u22121}{\\lim}f(x)[\/latex]. By looking at each one-sided limit as [latex]x\\to 1[\/latex], we see that<\/p>\r\n\r\n<div id=\"fs-id1165042491126\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 1^+}{\\lim}\\frac{x^2}{1-x^2}=\u2212\\infty[\/latex] and [latex]\\underset{x\\to 1^-}{\\lim}\\frac{x^2}{1-x^2}=\\infty[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165042491222\">In addition, by looking at each one-sided limit as [latex]x\\to \u22121[\/latex], we find that<\/p>\r\n\r\n<div id=\"fs-id1165042491242\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \u22121^+}{\\lim}\\frac{x^2}{1-x^2}=\\infty[\/latex] and [latex]\\underset{x\\to \u22121^-}{\\lim}\\frac{x^2}{1-x^2}=\u2212\\infty[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1165043262266\">Step 5. Calculate the first derivative:<\/p>\r\n\r\n<div id=\"fs-id1165043262270\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\frac{(1-x^2)(2x)-x^2(-2x)}{(1-x^2)^2}=\\frac{2x}{(1-x^2)^2}[\/latex].<\/div>\r\n<p id=\"fs-id1165043262390\">Critical points occur at points [latex]x[\/latex] where [latex]f^{\\prime}(x)=0[\/latex] or [latex]f^{\\prime}(x)[\/latex] is undefined. We see that [latex]f^{\\prime}(x)=0[\/latex] when [latex]x=0[\/latex]. The derivative [latex]f^{\\prime}[\/latex] is not undefined at any point in the domain of [latex]f[\/latex]. However, [latex]x=\\pm 1[\/latex] are not in the domain of [latex]f[\/latex]. Therefore, to determine where [latex]f[\/latex] is increasing and where [latex]f[\/latex] is decreasing, divide the interval [latex](\u2212\\infty ,\\infty )[\/latex] into four smaller intervals: [latex](\u2212\\infty ,-1)[\/latex], [latex](-1,0)[\/latex], [latex](0,1)[\/latex], and [latex](1,\\infty )[\/latex], and choose a test point in each interval to determine the sign of [latex]f^{\\prime}(x)[\/latex] in each of these intervals. The values [latex]x=-2[\/latex], [latex]x=-\\frac{1}{2}[\/latex], [latex]x=\\frac{1}{2}[\/latex], and [latex]x=2[\/latex] are good choices for test points as shown in the following table.<\/p>\r\n\r\n<table id=\"fs-id1165043341618\" class=\"unnumbered\" summary=\"This table has four columns and five rows. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019(x) = 2x\/(1 \u2212 x2)2, and Conclusion. Under the header row, the first column reads (\u2212\u221e, \u22121), (\u22121, 0), (0, 1), and (1, \u221e). The second column reads x = \u22122, x = \u22121\/2, x = 1\/2, and x = 2. The third column reads \u2212\/+ = \u2212, \u2212\/+ = \u2212, +\/+ = +, and +\/+ = +. The fourth column reads f is decreasing, f is decreasing, f is increasing, and f is increasing.\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Interval<\/th>\r\n<th>Test Point<\/th>\r\n<th>Sign of [latex]f^{\\prime}(x)=\\frac{2x}{(1-x^2)^2}[\/latex]<\/th>\r\n<th>Conclusion<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex](\u2212\\infty ,-1)[\/latex]<\/td>\r\n<td>[latex]x=-2[\/latex]<\/td>\r\n<td>[latex]\u2212\/+=\u2212[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is decreasing.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](-1,0)[\/latex]<\/td>\r\n<td>[latex]x=-1\/2[\/latex]<\/td>\r\n<td>[latex]\u2212\/+=\u2212[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is decreasing.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](0,1)[\/latex]<\/td>\r\n<td>[latex]x=1\/2[\/latex]<\/td>\r\n<td>[latex]+\/+=+[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is increasing.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](1,\\infty )[\/latex]<\/td>\r\n<td>[latex]x=2[\/latex]<\/td>\r\n<td>[latex]+\/+=+[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is increasing.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165043183348\">From this analysis, we conclude that [latex]f[\/latex] has a local minimum at [latex]x=0[\/latex] but no local maximum.<\/p>\r\n<p id=\"fs-id1165043183355\">Step 6. Calculate the second derivative:<\/p>\r\n\r\n<div id=\"fs-id1165043183358\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} f^{\\prime \\prime}(x) &amp; =\\frac{(1-x^2)^2(2)-2x(2(1-x^2)(-2x))}{(1-x^2)^4} \\\\ &amp; =\\frac{(1-x^2)[2(1-x^2)+8x^2]}{(1-x^2)^4} \\\\ &amp; =\\frac{2(1-x^2)+8x^2}{(1-x^2)^3} \\\\ &amp; =\\frac{6x^2+2}{(1-x^2)^3} \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165043384128\">To determine the intervals where [latex]f[\/latex] is concave up and where [latex]f[\/latex] is concave down, we first need to find all points [latex]x[\/latex] where [latex]f^{\\prime \\prime}(x)=0[\/latex] or [latex]f^{\\prime \\prime}(x)[\/latex] is undefined. Since the numerator [latex]6x^2+2 \\ne 0[\/latex] for any [latex]x[\/latex], [latex]f^{\\prime \\prime}(x)[\/latex] is never zero. Furthermore, [latex]f^{\\prime \\prime}[\/latex] is not undefined for any [latex]x[\/latex] in the domain of [latex]f[\/latex]. However, as discussed earlier, [latex]x=\\pm 1[\/latex] are not in the domain of [latex]f[\/latex]. Therefore, to determine the concavity of [latex]f[\/latex], we divide the interval [latex](\u2212\\infty ,\\infty )[\/latex] into the three smaller intervals [latex](\u2212\\infty ,-1)[\/latex], [latex](-1,-1)[\/latex], and [latex](1,\\infty )[\/latex], and choose a test point in each of these intervals to evaluate the sign of [latex]f^{\\prime \\prime}(x)[\/latex] in each of these intervals. The values [latex]x=-2[\/latex], [latex]x=0[\/latex], and [latex]x=2[\/latex] are possible test points as shown in the following table.<\/p>\r\n\r\n<table id=\"fs-id1165043384406\" class=\"unnumbered\" summary=\"This table has four columns and four rows. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019\u2019(x) = (6x2 + 2)\/(1 \u2212 x2)3, and Conclusion. Under the header row, the first column reads (\u2212\u221e, \u22121), (\u22121, 1), and (1, \u221e). The second column reads x = \u22122, x = 0, and x = 2. The third column reads +\/\u2212 = \u2212, +\/+ = +, and +\/\u2212 = \u2212. The fourth column reads f is concave down, f is concave up, and f is concave down.\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Interval<\/th>\r\n<th>Test Point<\/th>\r\n<th>Sign of [latex]f^{\\prime \\prime}(x)=\\frac{6x^2+2}{(1-x^2)^3}[\/latex]<\/th>\r\n<th>Conclusion<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex](\u2212\\infty ,-1)[\/latex]<\/td>\r\n<td>[latex]x=-2[\/latex]<\/td>\r\n<td>[latex]+\/-=\u2212[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is concave down.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](-1,-1)[\/latex]<\/td>\r\n<td>[latex]x=0[\/latex]<\/td>\r\n<td>[latex]+\/+=+[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is concave up.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](1,\\infty )[\/latex]<\/td>\r\n<td>[latex]x=2[\/latex]<\/td>\r\n<td>[latex]+\/-=\u2212[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is concave down.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165042592688\">Combining all this information, we arrive at the graph of [latex]f[\/latex] shown below. Note that, although [latex]f[\/latex] changes concavity at [latex]x=-1[\/latex] and [latex]x=1[\/latex], there are no inflection points at either of these places because [latex]f[\/latex] is not continuous at [latex]x=-1[\/latex] or [latex]x=1[\/latex].<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"342\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211139\/CNX_Calc_Figure_04_06_016.jpg\" alt=\"The function f(x) = x2\/(1 \u2212 x2) is graphed. It has asymptotes y = \u22121, x = \u22121, and x = 1.\" width=\"342\" height=\"347\" \/> Figure 25.[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042592764\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSketch a graph of [latex]f(x)=\\dfrac{3x+5}{8+4x}[\/latex]\r\n<div id=\"fs-id1165042592768\" class=\"exercise\">\r\n\r\n[reveal-answer q=\"3127001\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"3127001\"]\r\n<p id=\"fs-id1165042592850\">A line [latex]y=L[\/latex] is a horizontal asymptote of [latex]f[\/latex] if the limit as [latex]x\\to \\infty [\/latex] or the limit as [latex]x\\to \u2212\\infty [\/latex] of [latex]f(x)[\/latex] is [latex]L[\/latex]. A line [latex]x=a[\/latex] is a vertical asymptote if at least one of the one-sided limits of [latex]f[\/latex] as [latex]x\\to a[\/latex] is [latex]\\infty [\/latex] or [latex]\u2212\\infty[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165042592825\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042592825\"]\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"717\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211142\/CNX_Calc_Figure_04_06_029.jpg\" alt=\"The function f(x) = (3x + 5)\/(8 + 4x) is graphed. It appears to have asymptotes at x = \u22122 and y = 1.\" width=\"717\" height=\"422\" \/> Figure 26.[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042592955\" class=\"textbook exercises\">\r\n<h3>Example: Sketching a Rational Function with an Oblique Asymptote<\/h3>\r\nSketch the graph of [latex]f(x)=\\dfrac{x^2}{x-1}[\/latex]\r\n<div id=\"fs-id1165042592957\" class=\"exercise\">[reveal-answer q=\"fs-id1165042593006\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042593006\"]\r\n<p id=\"fs-id1165042593006\">Step 1. The domain of [latex]f[\/latex] is the set of all real numbers [latex]x[\/latex] except [latex]x=1[\/latex].<\/p>\r\n<p id=\"fs-id1165042712534\">Step 2. Find the intercepts. We can see that when [latex]x=0[\/latex], [latex]f(x)=0[\/latex], so [latex](0,0)[\/latex] is the only intercept.<\/p>\r\n<p id=\"fs-id1165042712584\">Step 3. Evaluate the limits at infinity. Since the degree of the numerator is one more than the degree of the denominator, [latex]f[\/latex] must have an oblique asymptote. To find the oblique asymptote, use long division of polynomials to write<\/p>\r\n\r\n<div id=\"fs-id1165042712594\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=\\frac{x^2}{x-1}=x+1+\\frac{1}{x-1}[\/latex].<\/div>\r\n<p id=\"fs-id1165042712649\">Since [latex]1\/(x-1)\\to 0[\/latex] as [latex]x\\to \\pm \\infty[\/latex], [latex]f(x)[\/latex] approaches the line [latex]y=x+1[\/latex] as [latex]x\\to \\pm \\infty[\/latex]. The line [latex]y=x+1[\/latex] is an oblique asymptote for [latex]f[\/latex].<\/p>\r\n<p id=\"fs-id1165042712758\">Step 4. To check for vertical asymptotes, look at where the denominator is zero. Here the denominator is zero at [latex]x=1[\/latex]. Looking at both one-sided limits as [latex]x\\to 1[\/latex], we find<\/p>\r\n\r\n<div id=\"fs-id1165042712789\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 1^+}{\\lim}\\frac{x^2}{x-1}=\\infty[\/latex] and [latex]\\underset{x\\to 1^-}{\\lim}\\frac{x^2}{x-1}=\u2212\\infty[\/latex].<\/div>\r\n<p id=\"fs-id1165042403315\">Therefore, [latex]x=1[\/latex] is a vertical asymptote, and we have determined the behavior of [latex]f[\/latex] as [latex]x[\/latex] approaches 1 from the right and the left.<\/p>\r\n<p id=\"fs-id1165042403340\">Step 5. Calculate the first derivative:<\/p>\r\n\r\n<div id=\"fs-id1165042403344\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\frac{(x-1)(2x)-x^2(1)}{(x-1)^2}=\\frac{x^2-2x}{(x-1)^2}[\/latex].<\/div>\r\n<p id=\"fs-id1165042403459\">We have [latex]f^{\\prime}(x)=0[\/latex] when [latex]x^2-2x=x(x-2)=0[\/latex]. Therefore, [latex]x=0[\/latex] and [latex]x=2[\/latex] are critical points. Since [latex]f[\/latex] is undefined at [latex]x=1[\/latex], we need to divide the interval [latex](\u2212\\infty ,\\infty )[\/latex] into the smaller intervals [latex](\u2212\\infty ,0)[\/latex], [latex](0,1)[\/latex], [latex](1,2)[\/latex], and [latex](2,\\infty )[\/latex], and choose a test point from each interval to evaluate the sign of [latex]f^{\\prime}(x)[\/latex] in each of these smaller intervals. For example, let [latex]x=-1[\/latex], [latex]x=\\frac{1}{2}[\/latex], [latex]x=\\frac{3}{2}[\/latex], and [latex]x=3[\/latex] be the test points as shown in the following table.<\/p>\r\n\r\n<table id=\"fs-id1165042710566\" class=\"unnumbered\" summary=\"This table has four columns and five rows. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019(x) = (x2 \u2212 2x)\/(x \u2212 1)2 = x(x \u2212 2)\/(x \u2212 1)2, and Conclusion. Under the header row, the first column reads (\u2212\u221e, 0), (0, 1), (1, 2), and (2, \u221e). The second column reads x = \u22121, x = 1\/2, x = 3\/2, and x = 3. The third column reads (\u2212)(\u2212)\/+ = +, (+)(\u2212)\/+ = \u2212, (+)(\u2212)\/+ = \u2212, and (+)(+)\/+ = +. The fourth column reads f is increasing, f is decreasing, f is decreasing, and f is increasing.\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Interval<\/th>\r\n<th>Test Point<\/th>\r\n<th>Sign of [latex]f^{\\prime}(x)=\\frac{x^2-2x}{(x-1)^2}=\\frac{x(x-2)}{(x-1)^2}[\/latex]<\/th>\r\n<th>Conclusion<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex](\u2212\\infty ,0)[\/latex]<\/td>\r\n<td>[latex]x=-1[\/latex]<\/td>\r\n<td>[latex](\u2212)(\u2212)\/+=+[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is increasing.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](0,1)[\/latex]<\/td>\r\n<td>[latex]x=1\/2[\/latex]<\/td>\r\n<td>[latex](+)(\u2212)\/+=\u2212[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is decreasing.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](1,2)[\/latex]<\/td>\r\n<td>[latex]x=3\/2[\/latex]<\/td>\r\n<td>[latex](+)(\u2212)\/+=\u2212[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is decreasing.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](2,\\infty )[\/latex]<\/td>\r\n<td>[latex]x=3[\/latex]<\/td>\r\n<td>[latex](+)(+)\/+=+[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is increasing.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165042476453\">From this table, we see that [latex]f[\/latex] has a local maximum at [latex]x=0[\/latex] and a local minimum at [latex]x=2[\/latex]. The value of [latex]f[\/latex] at the local maximum is [latex]f(0)=0[\/latex] and the value of [latex]f[\/latex] at the local minimum is [latex]f(2)=4[\/latex]. Therefore, [latex](0,0)[\/latex] and [latex](2,4)[\/latex] are important points on the graph.<\/p>\r\n<p id=\"fs-id1165042464546\">Step 6. Calculate the second derivative:<\/p>\r\n\r\n<div id=\"fs-id1165042464549\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} f^{\\prime \\prime}(x) &amp; =\\frac{(x-1)^2(2x-2)-(x^2-2x)(2(x-1))}{(x-1)^4} \\\\ &amp; =\\frac{(x-1)[(x-1)(2x-2)-2(x^2-2x)]}{(x-1)^4} \\\\ &amp; =\\frac{(x-1)(2x-2)-2(x^2-2x)}{(x-1)^3} \\\\ &amp; =\\frac{2x^2-4x+2-(2x^2-4x)}{(x-1)^3} \\\\ &amp; =\\frac{2}{(x-1)^3} \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165042652400\">We see that [latex]f^{\\prime \\prime}(x)[\/latex] is never zero or undefined for [latex]x[\/latex] in the domain of [latex]f[\/latex]. Since [latex]f[\/latex] is undefined at [latex]x=1[\/latex], to check concavity we just divide the interval [latex](\u2212\\infty ,\\infty )[\/latex] into the two smaller intervals [latex](\u2212\\infty ,1)[\/latex] and [latex](1,\\infty )[\/latex], and choose a test point from each interval to evaluate the sign of [latex]f^{\\prime \\prime}(x)[\/latex] in each of these intervals. The values [latex]x=0[\/latex] and [latex]x=2[\/latex] are possible test points as shown in the following table.<\/p>\r\n\r\n<table id=\"fs-id1165042652542\" class=\"unnumbered\" summary=\"This table has four columns and three rows. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019\u2019(x) = 2\/(x \u2212 1)3, and Conclusion. Under the header row, the first column reads (\u2212\u221e, 1) and (1, \u221e). The column row reads x = 0 and x = 2. The third column reads +\/\u2212 = \u2212 and +\/+ = +. The fourth column reads f is concave down and f is concave up.\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Interval<\/th>\r\n<th>Test Point<\/th>\r\n<th>Sign of [latex]f^{\\prime \\prime}(x)=\\frac{2}{(x-1)^3}[\/latex]<\/th>\r\n<th>Conclusion<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex](\u2212\\infty ,1)[\/latex]<\/td>\r\n<td>[latex]x=0[\/latex]<\/td>\r\n<td>[latex]+\/-=\u2212[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is concave down.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](1,\\infty )[\/latex]<\/td>\r\n<td>[latex]x=2[\/latex]<\/td>\r\n<td>[latex]+\/+=+[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is concave up.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165042606958\">From the information gathered, we arrive at the following graph for [latex]f.[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"342\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211145\/CNX_Calc_Figure_04_06_017.jpg\" alt=\"The function f(x) = x2\/(x \u2212 1) is graphed. It has asymptotes y = x + 1 and x = 1.\" width=\"342\" height=\"346\" \/> Figure 27.[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Sketching a Rational Function with an Oblique Asymptote.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/QhOY9VfIefo?controls=0&amp;start=552&amp;end=777&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266835\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266835\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.6LimitsAtInfinityAndAsymptotesPart2_552to777_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.6 Limits at Infinity and Asymptotes (part 2 - curve sketching)\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1165042606982\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the oblique asymptote for [latex]f(x)=\\dfrac{3x^3-2x+1}{2x^2-4}[\/latex]\r\n<div id=\"fs-id1165042606986\" class=\"exercise\">[reveal-answer q=\"99083451\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"99083451\"]<\/div>\r\n<div>\r\n<p id=\"fs-id1165042607084\">Use long division of polynomials.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"exercise\">\r\n<p id=\"fs-id1165042607059\">[reveal-answer q=\"fs-id1165042607059\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042607059\"]<\/p>\r\n[latex]y=\\frac{3}{2}x[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165042607090\" class=\"textbook exercises\">\r\n<h3>Example: Sketching the Graph of a Function with a Cusp<\/h3>\r\nSketch a graph of [latex]f(x)=(x-1)^{\\frac{2}{3}}[\/latex]\r\n<div id=\"fs-id1165042607093\" class=\"exercise\">[reveal-answer q=\"fs-id1165042607145\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042607145\"]\r\n<p id=\"fs-id1165042607145\">Step 1. Since the cube-root function is defined for all real numbers [latex]x[\/latex] and [latex](x-1)^{2\/3}=(\\sqrt[3]{x-1})^2[\/latex], the domain of [latex]f[\/latex] is all real numbers.<\/p>\r\n<p id=\"fs-id1165042607208\">Step 2: To find the [latex]y[\/latex]-intercept, evaluate [latex]f(0)[\/latex]. Since [latex]f(0)=1[\/latex], the [latex]y[\/latex]-intercept is [latex](0,1)[\/latex]. To find the [latex]x[\/latex]-intercept, solve [latex](x-1)^{2\/3}=0[\/latex]. The solution of this equation is [latex]x=1[\/latex], so the [latex]x[\/latex]-intercept is [latex](1,0)[\/latex].<\/p>\r\n<p id=\"fs-id1165042583208\">Step 3: Since [latex]\\underset{x\\to \\pm \\infty }{\\lim}(x-1)^{2\/3}=\\infty [\/latex], the function continues to grow without bound as [latex]x\\to \\infty [\/latex] and [latex]x\\to \u2212\\infty[\/latex].<\/p>\r\n<p id=\"fs-id1165042583285\">Step 4: The function has no vertical asymptotes.<\/p>\r\n<p id=\"fs-id1165042583288\">Step 5: To determine where [latex]f[\/latex] is increasing or decreasing, calculate [latex]f^{\\prime}[\/latex]. We find<\/p>\r\n\r\n<div id=\"fs-id1165042583307\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\frac{2}{3}(x-1)^{-1\/3}=\\frac{2}{3(x-1)^{1\/3}}[\/latex].<\/div>\r\n<p id=\"fs-id1165042583388\">This function is not zero anywhere, but it is undefined when [latex]x=1[\/latex]. Therefore, the only critical point is [latex]x=1[\/latex]. Divide the interval [latex](\u2212\\infty ,\\infty )[\/latex] into the smaller intervals [latex](\u2212\\infty ,1)[\/latex] and [latex](1,\\infty )[\/latex], and choose test points in each of these intervals to determine the sign of [latex]f^{\\prime}(x)[\/latex] in each of these smaller intervals. Let [latex]x=0[\/latex] and [latex]x=2[\/latex] be the test points as shown in the following table.<\/p>\r\n\r\n<table id=\"fs-id1165042504467\" class=\"unnumbered\" summary=\"This table has four columns and three rows. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019(x) = 2\/(3(x \u2212 1)1\/3), and Conclusion. Under the header row, the first column reads (\u2212\u221e, 1) and (1, \u221e). The second column reads x = 0 and x = 2. The third column reads +\/\u2212 = \u2212 and +\/+ = +. The fourth column reads f is decreasing and f is increasing.\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Interval<\/th>\r\n<th>Test Point<\/th>\r\n<th>Sign of [latex]f^{\\prime}(x)=\\frac{2}{3(x-1)^{1\/3}}[\/latex]<\/th>\r\n<th>Conclusion<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex](\u2212\\infty ,1)[\/latex]<\/td>\r\n<td>[latex]x=0[\/latex]<\/td>\r\n<td>[latex]+\/-=\u2212[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is decreasing.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](1,\\infty )[\/latex]<\/td>\r\n<td>[latex]x=2[\/latex]<\/td>\r\n<td>[latex]+\/+=+[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is increasing.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165042504734\">We conclude that [latex]f[\/latex] has a local minimum at [latex]x=1[\/latex]. Evaluating [latex]f[\/latex] at [latex]x=1[\/latex], we find that the value of [latex]f[\/latex] at the local minimum is zero. Note that [latex]f^{\\prime}(1)[\/latex] is undefined, so to determine the behavior of the function at this critical point, we need to examine [latex]\\underset{x\\to 1}{\\lim}f^{\\prime}(x)[\/latex]. Looking at the one-sided limits, we have<\/p>\r\n\r\n<div id=\"fs-id1165042510168\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 1^+}{\\lim}\\frac{2}{3(x-1)^{1\/3}}=\\infty[\/latex] and [latex]\\underset{x\\to 1^-}{\\lim}\\frac{2}{3(x-1)^{1\/3}}=\u2212\\infty[\/latex].<\/div>\r\n<p id=\"fs-id1165042510285\">Therefore, [latex]f[\/latex] has a cusp at [latex]x=1[\/latex].<\/p>\r\n<p id=\"fs-id1165042510304\">Step 6: To determine concavity, we calculate the second derivative of [latex]f[\/latex]:<\/p>\r\n\r\n<div id=\"fs-id1165042510313\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime \\prime}(x)=-\\frac{2}{9}(x-1)^{-4\/3}=\\frac{-2}{9(x-1)^{4\/3}}[\/latex].<\/div>\r\n<p id=\"fs-id1165042510398\">We find that [latex]f^{\\prime \\prime}(x)[\/latex] is defined for all [latex]x[\/latex], but is undefined when [latex]x=1[\/latex]. Therefore, divide the interval [latex](\u2212\\infty ,\\infty )[\/latex] into the smaller intervals [latex](\u2212\\infty ,1)[\/latex] and [latex](1,\\infty )[\/latex], and choose test points to evaluate the sign of [latex]f^{\\prime \\prime}(x)[\/latex] in each of these intervals. As we did earlier, let [latex]x=0[\/latex] and [latex]x=2[\/latex] be test points as shown in the following table.<\/p>\r\n\r\n<table id=\"fs-id1165042510530\" class=\"unnumbered\" summary=\"This table has four columns and three rows. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019\u2019(x) = \u22122\/(9(x \u2212 1)4\/3), and Conclusion. Under the header row, the first column reads (\u2212\u221e, 1) and (1, \u221e). The second column reads x = 0 and x = 2. The third column reads \u2212\/+ = \u2212 and \u2212\/+ = \u2212. The fourth column reads f is concave down and f is concave down.\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Interval<\/th>\r\n<th>Test Point<\/th>\r\n<th>Sign of [latex]f^{\\prime \\prime}(x)=\\frac{-2}{9(x-1)^{4\/3}}[\/latex]<\/th>\r\n<th>Conclusion<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex](\u2212\\infty ,1)[\/latex]<\/td>\r\n<td>[latex]x=0[\/latex]<\/td>\r\n<td>[latex]\u2212\/+=\u2212[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is concave down.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](1,\\infty )[\/latex]<\/td>\r\n<td>[latex]x=2[\/latex]<\/td>\r\n<td>[latex]\u2212\/+=\u2212[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is concave down.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165042643969\">From this table, we conclude that [latex]f[\/latex] is concave down everywhere. Combining all of this information, we arrive at the following graph for [latex]f[\/latex].<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"604\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211148\/CNX_Calc_Figure_04_06_018.jpg\" alt=\"The function f(x) = (x \u2212 1)2\/3 is graphed. It touches the x axis at x = 1, where it comes to something of a sharp point and then flairs out on either side.\" width=\"604\" height=\"272\" \/> Figure 28.[\/caption]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Sketching the Graph of a Function with a Cusp.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/QhOY9VfIefo?controls=0&amp;start=781&amp;end=962&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.6LimitsAtInfinityAndAsymptotesPart2_781to962_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.6 Limits at Infinity and Asymptotes (part 2 - curve sketching)\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nConsider the function [latex]f(x)=5-x^{\\frac{2}{3}}[\/latex]. Determine the point on the graph where a cusp is located. Determine the end behavior of [latex]f[\/latex].\r\n\r\n[reveal-answer q=\"44668822\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"44668822\"]\r\n\r\nA function [latex]f[\/latex] has a cusp at a point [latex]a[\/latex] if [latex]f(a)[\/latex] exists, [latex]f^{\\prime}(a)[\/latex] is undefined, one of the one-sided limits as [latex]x\\to a[\/latex] of [latex]f^{\\prime}(x)[\/latex] is [latex]+\\infty[\/latex], and the other one-sided limit is [latex]\u2212\\infty[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1165042644060\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1165042644060\"]\r\n\r\nThe function [latex]f[\/latex] has a cusp at [latex](0,5)[\/latex]: [latex]\\underset{x\\to 0^-}{\\lim}f^{\\prime}(x)=\\infty[\/latex], [latex]\\underset{x\\to 0^+}{\\lim}f^{\\prime}(x)=\u2212\\infty[\/latex]. For end behavior, [latex]\\underset{x\\to \\pm \\infty }{\\lim}f(x)=\u2212\\infty[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Analyze a function and its derivatives to draw its graph<\/li>\n<\/ul>\n<\/div>\n<h3>Guidelines for Graphing a Function<\/h3>\n<p id=\"fs-id1165042602938\">We now have enough analytical tools to draw graphs of a wide variety of algebraic and transcendental functions. Before showing how to graph specific functions, let\u2019s look at a general strategy to use when graphing any function.<\/p>\n<div id=\"fs-id1165042602944\" class=\"textbox examples\">\n<h3>Problem-Solving Strategy: Drawing the Graph of a Function<\/h3>\n<p id=\"fs-id1165042602951\">Given a function [latex]f[\/latex] use the following steps to sketch a graph of [latex]f[\/latex]:<\/p>\n<ol id=\"fs-id1165042602969\">\n<li>Determine the domain of the function.<\/li>\n<li>Locate the [latex]x[\/latex]&#8211; and [latex]y[\/latex]-intercepts.<\/li>\n<li>Evaluate [latex]\\underset{x\\to \\infty }{\\lim}f(x)[\/latex] and [latex]\\underset{x\\to \u2212\\infty }{\\lim}f(x)[\/latex] to determine the end behavior. If either of these limits is a finite number [latex]L[\/latex], then [latex]y=L[\/latex] is a horizontal asymptote. If either of these limits is [latex]\\infty[\/latex] or [latex]\u2212\\infty[\/latex], determine whether [latex]f[\/latex] has an oblique asymptote. If [latex]f[\/latex] is a rational function such that [latex]f(x)=\\frac{p(x)}{q(x)}[\/latex], where the degree of the numerator is greater than the degree of the denominator, then [latex]f[\/latex] can be written as\n<div id=\"fs-id1165042617521\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=\\dfrac{p(x)}{q(x)}=g(x)+\\dfrac{r(x)}{q(x)}[\/latex],<\/div>\n<p>where the degree of [latex]r(x)[\/latex] is less than the degree of [latex]q(x)[\/latex]. The values of [latex]f(x)[\/latex] approach the values of [latex]g(x)[\/latex] as [latex]x\\to \\pm \\infty[\/latex]. If [latex]g(x)[\/latex] is a linear function, it is known as an <em>oblique asymptote<\/em>.<\/li>\n<li>Determine whether [latex]f[\/latex] has any vertical asymptotes.<\/li>\n<li>Calculate [latex]f^{\\prime}[\/latex]. Find all critical points and determine the intervals where [latex]f[\/latex] is increasing and where [latex]f[\/latex] is decreasing. Determine whether [latex]f[\/latex] has any local extrema.<\/li>\n<li>Calculate [latex]f^{\\prime \\prime}[\/latex]. Determine the intervals where [latex]f[\/latex] is concave up and where [latex]f[\/latex] is concave down. Use this information to determine whether [latex]f[\/latex] has any inflection points. The second derivative can also be used as an alternate means to determine or verify that [latex]f[\/latex] has a local extremum at a critical point.<\/li>\n<\/ol>\n<\/div>\n<p id=\"fs-id1165042617762\">Now let\u2019s use this strategy to graph several different functions. We start by graphing a polynomial function.<\/p>\n<div id=\"fs-id1165042617768\" class=\"textbook exercises\">\n<h3>Example: Sketching a Graph of a Polynomial<\/h3>\n<p>Sketch a graph of [latex]f(x)=(x-1)^2 (x+2)[\/latex]<\/p>\n<div id=\"fs-id1165042617770\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042707761\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042707761\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042707761\">Step 1. Since [latex]f[\/latex] is a polynomial, the domain is the set of all real numbers.<\/p>\n<p id=\"fs-id1165042707768\">Step 2. When [latex]x=0, \\, f(x)=2[\/latex]. Therefore, the [latex]y[\/latex]-intercept is [latex](0,2)[\/latex]. To find the [latex]x[\/latex]-intercepts, we need to solve the equation [latex](x-1)^2 (x+2)=0[\/latex], which gives us the [latex]x[\/latex]-intercepts [latex](1,0)[\/latex] and [latex](-2,0)[\/latex]<\/p>\n<p id=\"fs-id1165042707901\">Step 3. We need to evaluate the end behavior of [latex]f[\/latex]. As [latex]x\\to \\infty[\/latex], [latex](x-1)^2 \\to \\infty[\/latex] and [latex](x+2)\\to \\infty[\/latex]. Therefore, [latex]\\underset{x\\to \\infty }{\\lim}f(x)=\\infty[\/latex]. As [latex]x\\to \u2212\\infty[\/latex], [latex](x-1)^2 \\to \\infty[\/latex] and [latex](x+2) \\to \u2212\\infty[\/latex]. Therefore, [latex]\\underset{x\\to -\\infty }{\\lim}f(x)=\u2212\\infty[\/latex]. To get even more information about the end behavior of [latex]f[\/latex], we can multiply the factors of [latex]f[\/latex]. When doing so, we see that<\/p>\n<div id=\"fs-id1165042525463\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=(x-1)^2 (x+2)=x^3-3x+2[\/latex]<\/div>\n<p id=\"fs-id1165042525532\">Since the leading term of [latex]f[\/latex] is [latex]x^3[\/latex], we conclude that [latex]f[\/latex] behaves like [latex]y=x^3[\/latex] as [latex]x\\to \\pm \\infty[\/latex].<\/p>\n<p id=\"fs-id1165043382921\">Step 4. Since [latex]f[\/latex] is a polynomial function, it does not have any vertical asymptotes.<\/p>\n<p id=\"fs-id1165043382928\">Step 5. The first derivative of [latex]f[\/latex] is<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=3x^2-3[\/latex]<\/div>\n<p id=\"fs-id1165043382970\">Therefore, [latex]f[\/latex] has two critical points: [latex]x=1,-1[\/latex]. Divide the interval [latex](\u2212\\infty ,\\infty)[\/latex] into the three smaller intervals: [latex](\u2212\\infty ,-1)[\/latex], [latex](-1,1)[\/latex], and [latex](1,\\infty )[\/latex]. Then, choose test points [latex]x=-2[\/latex], [latex]x=0[\/latex], and [latex]x=2[\/latex] from these intervals and evaluate the sign of [latex]f^{\\prime}(x)[\/latex] at each of these test points, as shown in the following table.<\/p>\n<table id=\"fs-id1165043383127\" class=\"unnumbered\" summary=\"This table has four rows and four columns. The first row is a header row, and it reads Interval, Test Point, Sign of Derivative f\u2019(x) = 3x2 \u2013 3 = 3(x \u2013 1)(x + 1), and Conclusion. Under the header row, the first column reads (\u2212\u221e, \u22121), (\u22121, 1), and (1, \u221e). The second column reads x = \u22122, x = 0, and x = 2. The third column reads (+)(\u2212)(\u2212) = +, (+)(\u2212)(+) = \u2212, and (+)(+)(+) = +. The fourth column reads f is increasing, f is decreasing, and f is increasing.\">\n<thead>\n<tr valign=\"top\">\n<th>Interval<\/th>\n<th>Test Point<\/th>\n<th>Sign of Derivative [latex]f^{\\prime}(x)=3x^2-3=3(x-1)(x+1)[\/latex]<\/th>\n<th>Conclusion<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex](\u2212\\infty ,-1)[\/latex]<\/td>\n<td>[latex]x=-2[\/latex]<\/td>\n<td>[latex](+)(\u2212)(\u2212)=+[\/latex]<\/td>\n<td>[latex]f[\/latex] is increasing.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](-1,1)[\/latex]<\/td>\n<td>[latex]x=0[\/latex]<\/td>\n<td>[latex](+)(\u2212)(+)=\u2212[\/latex]<\/td>\n<td>[latex]f[\/latex] is decreasing.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](1,\\infty )[\/latex]<\/td>\n<td>[latex]x=2[\/latex]<\/td>\n<td>[latex](+)(+)(+)=+[\/latex]<\/td>\n<td>[latex]f[\/latex] is increasing.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165042539204\">From the table, we see that [latex]f[\/latex] has a local maximum at [latex]x=-1[\/latex] and a local minimum at [latex]x=1[\/latex]. Evaluating [latex]f(x)[\/latex] at those two points, we find that the local maximum value is [latex]f(-1)=4[\/latex] and the local minimum value is [latex]f(1)=0[\/latex].<\/p>\n<p id=\"fs-id1165042539284\">Step 6. The second derivative of [latex]f[\/latex] is<\/p>\n<div id=\"fs-id1165042539292\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime \\prime}(x)=6x[\/latex]<\/div>\n<p id=\"fs-id1165042539318\">The second derivative is zero at [latex]x=0[\/latex]. Therefore, to determine the concavity of [latex]f[\/latex], divide the interval [latex](\u2212\\infty ,\\infty)[\/latex] into the smaller intervals [latex](\u2212\\infty ,0)[\/latex] and [latex](0,\\infty )[\/latex], and choose test points [latex]x=-1[\/latex] and [latex]x=1[\/latex] to determine the concavity of [latex]f[\/latex] on each of these smaller intervals as shown in the following table.<\/p>\n<table id=\"fs-id1165042381927\" class=\"unnumbered\" summary=\"This table has three rows and four columns. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019\u2019(x) = 6x, and Conclusion. Under the header row, the first column reads (\u2212\u221e, 0) and (0, \u221e). The second column reads x = \u22121 and x = 1. The third column reads \u2212 and +. The fourth column reads f is concave down and f is concave up.\">\n<thead>\n<tr valign=\"top\">\n<th>Interval<\/th>\n<th>Test Point<\/th>\n<th>Sign of [latex]f^{\\prime \\prime}(x)=6x[\/latex]<\/th>\n<th>Conclusion<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex](\u2212\\infty ,0)[\/latex]<\/td>\n<td>[latex]x=-1[\/latex]<\/td>\n<td>[latex]-[\/latex]<\/td>\n<td>[latex]f[\/latex] is concave down.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](0,\\infty )[\/latex]<\/td>\n<td>[latex]x=1[\/latex]<\/td>\n<td>[latex]+[\/latex]<\/td>\n<td>[latex]f[\/latex] is concave up.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165042382153\">We note that the information in the preceding table confirms the fact, found in step 5, that [latex]f[\/latex] has a local maximum at [latex]x=-1[\/latex] and a local minimum at [latex]x=1[\/latex]. In addition, the information found in step 5\u2014namely, [latex]f[\/latex] has a local maximum at [latex]x=-1[\/latex] and a local minimum at [latex]x=1[\/latex], and [latex]f^{\\prime}(x)=0[\/latex] at those points\u2014combined with the fact that [latex]f^{\\prime \\prime}[\/latex] changes sign only at [latex]x=0[\/latex] confirms the results found in step 6 on the concavity of [latex]f[\/latex].<\/p>\n<p id=\"fs-id1165042709981\">Combining this information, we arrive at the graph of [latex]f(x)=(x-1)^2 (x+2)[\/latex] shown in the following graph.<\/p>\n<div style=\"width: 410px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211132\/CNX_Calc_Figure_04_06_015.jpg\" alt=\"The function f(x) = (x \u22121)2 (x + 2) is graphed. It crosses the x axis at x = \u22122 and touches the x axis at x = 1.\" width=\"400\" height=\"347\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 23.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Sketching a Graph of a Polynomial.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/QhOY9VfIefo?controls=0&amp;start=003&amp;end=317&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.6LimitsAtInfinityAndAsymptotesPart2_3to317_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.6 Limits at Infinity and Asymptotes (part 2 &#8211; curve sketching)&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1165042710046\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Sketch a graph of [latex]f(x)=(x-1)^3 (x+2)[\/latex]<\/p>\n<div id=\"fs-id1165042710050\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q80046723\">Hint<\/span><\/p>\n<div id=\"q80046723\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042710130\">[latex]f[\/latex] is a fourth-degree polynomial.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042710106\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042710106\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"width: 411px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211136\/CNX_Calc_Figure_04_06_028.jpg\" alt=\"The function f(x) = (x \u22121)3(x + 2) is graphed.\" width=\"401\" height=\"520\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 24.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042710140\" class=\"textbook exercises\">\n<h3>Example: Sketching a Rational Function<\/h3>\n<p>Sketch the graph of [latex]f(x)=\\dfrac{x^2}{1-x^2}[\/latex]<\/p>\n<div id=\"fs-id1165042710142\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042407414\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042407414\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042407414\">Step 1. The function [latex]f[\/latex] is defined as long as the denominator is not zero. Therefore, the domain is the set of all real numbers [latex]x[\/latex] except [latex]x=\\pm 1[\/latex].<\/p>\n<p id=\"fs-id1165042407440\">Step 2. Find the intercepts. If [latex]x=0[\/latex], then [latex]f(x)=0[\/latex], so 0 is an intercept. If [latex]y=0[\/latex], then [latex]\\frac{x^2}{1-x^2}=0[\/latex], which implies [latex]x=0[\/latex]. Therefore, [latex](0,0)[\/latex] is the only intercept.<\/p>\n<p id=\"fs-id1165042407553\">Step 3. Evaluate the limits at infinity. Since [latex]f[\/latex] is a rational function, divide the numerator and denominator by the highest power in the denominator: [latex]x^2[\/latex]. We obtain<\/p>\n<div id=\"fs-id1165042407571\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\pm \\infty }{\\lim}\\frac{x^2}{1-x^2}=\\underset{x\\to \\pm \\infty }{\\lim}\\frac{1}{\\frac{1}{x^2}-1}=-1[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042407655\">Therefore, [latex]f[\/latex] has a horizontal asymptote of [latex]y=-1[\/latex] as [latex]x\\to \\infty[\/latex] and [latex]x\\to \u2212\\infty[\/latex].<\/p>\n<p id=\"fs-id1165042491004\">Step 4. To determine whether [latex]f[\/latex] has any vertical asymptotes, first check to see whether the denominator has any zeroes. We find the denominator is zero when [latex]x=\\pm 1[\/latex]. To determine whether the lines [latex]x=1[\/latex] or [latex]x=-1[\/latex] are vertical asymptotes of [latex]f[\/latex], evaluate [latex]\\underset{x\\to 1}{\\lim}f(x)[\/latex] and [latex]\\underset{x\\to \u22121}{\\lim}f(x)[\/latex]. By looking at each one-sided limit as [latex]x\\to 1[\/latex], we see that<\/p>\n<div id=\"fs-id1165042491126\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 1^+}{\\lim}\\frac{x^2}{1-x^2}=\u2212\\infty[\/latex] and [latex]\\underset{x\\to 1^-}{\\lim}\\frac{x^2}{1-x^2}=\\infty[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042491222\">In addition, by looking at each one-sided limit as [latex]x\\to \u22121[\/latex], we find that<\/p>\n<div id=\"fs-id1165042491242\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \u22121^+}{\\lim}\\frac{x^2}{1-x^2}=\\infty[\/latex] and [latex]\\underset{x\\to \u22121^-}{\\lim}\\frac{x^2}{1-x^2}=\u2212\\infty[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165043262266\">Step 5. Calculate the first derivative:<\/p>\n<div id=\"fs-id1165043262270\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\frac{(1-x^2)(2x)-x^2(-2x)}{(1-x^2)^2}=\\frac{2x}{(1-x^2)^2}[\/latex].<\/div>\n<p id=\"fs-id1165043262390\">Critical points occur at points [latex]x[\/latex] where [latex]f^{\\prime}(x)=0[\/latex] or [latex]f^{\\prime}(x)[\/latex] is undefined. We see that [latex]f^{\\prime}(x)=0[\/latex] when [latex]x=0[\/latex]. The derivative [latex]f^{\\prime}[\/latex] is not undefined at any point in the domain of [latex]f[\/latex]. However, [latex]x=\\pm 1[\/latex] are not in the domain of [latex]f[\/latex]. Therefore, to determine where [latex]f[\/latex] is increasing and where [latex]f[\/latex] is decreasing, divide the interval [latex](\u2212\\infty ,\\infty )[\/latex] into four smaller intervals: [latex](\u2212\\infty ,-1)[\/latex], [latex](-1,0)[\/latex], [latex](0,1)[\/latex], and [latex](1,\\infty )[\/latex], and choose a test point in each interval to determine the sign of [latex]f^{\\prime}(x)[\/latex] in each of these intervals. The values [latex]x=-2[\/latex], [latex]x=-\\frac{1}{2}[\/latex], [latex]x=\\frac{1}{2}[\/latex], and [latex]x=2[\/latex] are good choices for test points as shown in the following table.<\/p>\n<table id=\"fs-id1165043341618\" class=\"unnumbered\" summary=\"This table has four columns and five rows. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019(x) = 2x\/(1 \u2212 x2)2, and Conclusion. Under the header row, the first column reads (\u2212\u221e, \u22121), (\u22121, 0), (0, 1), and (1, \u221e). The second column reads x = \u22122, x = \u22121\/2, x = 1\/2, and x = 2. The third column reads \u2212\/+ = \u2212, \u2212\/+ = \u2212, +\/+ = +, and +\/+ = +. The fourth column reads f is decreasing, f is decreasing, f is increasing, and f is increasing.\">\n<thead>\n<tr valign=\"top\">\n<th>Interval<\/th>\n<th>Test Point<\/th>\n<th>Sign of [latex]f^{\\prime}(x)=\\frac{2x}{(1-x^2)^2}[\/latex]<\/th>\n<th>Conclusion<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex](\u2212\\infty ,-1)[\/latex]<\/td>\n<td>[latex]x=-2[\/latex]<\/td>\n<td>[latex]\u2212\/+=\u2212[\/latex]<\/td>\n<td>[latex]f[\/latex] is decreasing.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](-1,0)[\/latex]<\/td>\n<td>[latex]x=-1\/2[\/latex]<\/td>\n<td>[latex]\u2212\/+=\u2212[\/latex]<\/td>\n<td>[latex]f[\/latex] is decreasing.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](0,1)[\/latex]<\/td>\n<td>[latex]x=1\/2[\/latex]<\/td>\n<td>[latex]+\/+=+[\/latex]<\/td>\n<td>[latex]f[\/latex] is increasing.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](1,\\infty )[\/latex]<\/td>\n<td>[latex]x=2[\/latex]<\/td>\n<td>[latex]+\/+=+[\/latex]<\/td>\n<td>[latex]f[\/latex] is increasing.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165043183348\">From this analysis, we conclude that [latex]f[\/latex] has a local minimum at [latex]x=0[\/latex] but no local maximum.<\/p>\n<p id=\"fs-id1165043183355\">Step 6. Calculate the second derivative:<\/p>\n<div id=\"fs-id1165043183358\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} f^{\\prime \\prime}(x) & =\\frac{(1-x^2)^2(2)-2x(2(1-x^2)(-2x))}{(1-x^2)^4} \\\\ & =\\frac{(1-x^2)[2(1-x^2)+8x^2]}{(1-x^2)^4} \\\\ & =\\frac{2(1-x^2)+8x^2}{(1-x^2)^3} \\\\ & =\\frac{6x^2+2}{(1-x^2)^3} \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165043384128\">To determine the intervals where [latex]f[\/latex] is concave up and where [latex]f[\/latex] is concave down, we first need to find all points [latex]x[\/latex] where [latex]f^{\\prime \\prime}(x)=0[\/latex] or [latex]f^{\\prime \\prime}(x)[\/latex] is undefined. Since the numerator [latex]6x^2+2 \\ne 0[\/latex] for any [latex]x[\/latex], [latex]f^{\\prime \\prime}(x)[\/latex] is never zero. Furthermore, [latex]f^{\\prime \\prime}[\/latex] is not undefined for any [latex]x[\/latex] in the domain of [latex]f[\/latex]. However, as discussed earlier, [latex]x=\\pm 1[\/latex] are not in the domain of [latex]f[\/latex]. Therefore, to determine the concavity of [latex]f[\/latex], we divide the interval [latex](\u2212\\infty ,\\infty )[\/latex] into the three smaller intervals [latex](\u2212\\infty ,-1)[\/latex], [latex](-1,-1)[\/latex], and [latex](1,\\infty )[\/latex], and choose a test point in each of these intervals to evaluate the sign of [latex]f^{\\prime \\prime}(x)[\/latex] in each of these intervals. The values [latex]x=-2[\/latex], [latex]x=0[\/latex], and [latex]x=2[\/latex] are possible test points as shown in the following table.<\/p>\n<table id=\"fs-id1165043384406\" class=\"unnumbered\" summary=\"This table has four columns and four rows. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019\u2019(x) = (6x2 + 2)\/(1 \u2212 x2)3, and Conclusion. Under the header row, the first column reads (\u2212\u221e, \u22121), (\u22121, 1), and (1, \u221e). The second column reads x = \u22122, x = 0, and x = 2. The third column reads +\/\u2212 = \u2212, +\/+ = +, and +\/\u2212 = \u2212. The fourth column reads f is concave down, f is concave up, and f is concave down.\">\n<thead>\n<tr valign=\"top\">\n<th>Interval<\/th>\n<th>Test Point<\/th>\n<th>Sign of [latex]f^{\\prime \\prime}(x)=\\frac{6x^2+2}{(1-x^2)^3}[\/latex]<\/th>\n<th>Conclusion<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex](\u2212\\infty ,-1)[\/latex]<\/td>\n<td>[latex]x=-2[\/latex]<\/td>\n<td>[latex]+\/-=\u2212[\/latex]<\/td>\n<td>[latex]f[\/latex] is concave down.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](-1,-1)[\/latex]<\/td>\n<td>[latex]x=0[\/latex]<\/td>\n<td>[latex]+\/+=+[\/latex]<\/td>\n<td>[latex]f[\/latex] is concave up.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](1,\\infty )[\/latex]<\/td>\n<td>[latex]x=2[\/latex]<\/td>\n<td>[latex]+\/-=\u2212[\/latex]<\/td>\n<td>[latex]f[\/latex] is concave down.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165042592688\">Combining all this information, we arrive at the graph of [latex]f[\/latex] shown below. Note that, although [latex]f[\/latex] changes concavity at [latex]x=-1[\/latex] and [latex]x=1[\/latex], there are no inflection points at either of these places because [latex]f[\/latex] is not continuous at [latex]x=-1[\/latex] or [latex]x=1[\/latex].<\/p>\n<div style=\"width: 352px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211139\/CNX_Calc_Figure_04_06_016.jpg\" alt=\"The function f(x) = x2\/(1 \u2212 x2) is graphed. It has asymptotes y = \u22121, x = \u22121, and x = 1.\" width=\"342\" height=\"347\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 25.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042592764\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Sketch a graph of [latex]f(x)=\\dfrac{3x+5}{8+4x}[\/latex]<\/p>\n<div id=\"fs-id1165042592768\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q3127001\">Hint<\/span><\/p>\n<div id=\"q3127001\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042592850\">A line [latex]y=L[\/latex] is a horizontal asymptote of [latex]f[\/latex] if the limit as [latex]x\\to \\infty[\/latex] or the limit as [latex]x\\to \u2212\\infty[\/latex] of [latex]f(x)[\/latex] is [latex]L[\/latex]. A line [latex]x=a[\/latex] is a vertical asymptote if at least one of the one-sided limits of [latex]f[\/latex] as [latex]x\\to a[\/latex] is [latex]\\infty[\/latex] or [latex]\u2212\\infty[\/latex].<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042592825\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042592825\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"width: 727px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211142\/CNX_Calc_Figure_04_06_029.jpg\" alt=\"The function f(x) = (3x + 5)\/(8 + 4x) is graphed. It appears to have asymptotes at x = \u22122 and y = 1.\" width=\"717\" height=\"422\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 26.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042592955\" class=\"textbook exercises\">\n<h3>Example: Sketching a Rational Function with an Oblique Asymptote<\/h3>\n<p>Sketch the graph of [latex]f(x)=\\dfrac{x^2}{x-1}[\/latex]<\/p>\n<div id=\"fs-id1165042592957\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042593006\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042593006\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042593006\">Step 1. The domain of [latex]f[\/latex] is the set of all real numbers [latex]x[\/latex] except [latex]x=1[\/latex].<\/p>\n<p id=\"fs-id1165042712534\">Step 2. Find the intercepts. We can see that when [latex]x=0[\/latex], [latex]f(x)=0[\/latex], so [latex](0,0)[\/latex] is the only intercept.<\/p>\n<p id=\"fs-id1165042712584\">Step 3. Evaluate the limits at infinity. Since the degree of the numerator is one more than the degree of the denominator, [latex]f[\/latex] must have an oblique asymptote. To find the oblique asymptote, use long division of polynomials to write<\/p>\n<div id=\"fs-id1165042712594\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=\\frac{x^2}{x-1}=x+1+\\frac{1}{x-1}[\/latex].<\/div>\n<p id=\"fs-id1165042712649\">Since [latex]1\/(x-1)\\to 0[\/latex] as [latex]x\\to \\pm \\infty[\/latex], [latex]f(x)[\/latex] approaches the line [latex]y=x+1[\/latex] as [latex]x\\to \\pm \\infty[\/latex]. The line [latex]y=x+1[\/latex] is an oblique asymptote for [latex]f[\/latex].<\/p>\n<p id=\"fs-id1165042712758\">Step 4. To check for vertical asymptotes, look at where the denominator is zero. Here the denominator is zero at [latex]x=1[\/latex]. Looking at both one-sided limits as [latex]x\\to 1[\/latex], we find<\/p>\n<div id=\"fs-id1165042712789\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 1^+}{\\lim}\\frac{x^2}{x-1}=\\infty[\/latex] and [latex]\\underset{x\\to 1^-}{\\lim}\\frac{x^2}{x-1}=\u2212\\infty[\/latex].<\/div>\n<p id=\"fs-id1165042403315\">Therefore, [latex]x=1[\/latex] is a vertical asymptote, and we have determined the behavior of [latex]f[\/latex] as [latex]x[\/latex] approaches 1 from the right and the left.<\/p>\n<p id=\"fs-id1165042403340\">Step 5. Calculate the first derivative:<\/p>\n<div id=\"fs-id1165042403344\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\frac{(x-1)(2x)-x^2(1)}{(x-1)^2}=\\frac{x^2-2x}{(x-1)^2}[\/latex].<\/div>\n<p id=\"fs-id1165042403459\">We have [latex]f^{\\prime}(x)=0[\/latex] when [latex]x^2-2x=x(x-2)=0[\/latex]. Therefore, [latex]x=0[\/latex] and [latex]x=2[\/latex] are critical points. Since [latex]f[\/latex] is undefined at [latex]x=1[\/latex], we need to divide the interval [latex](\u2212\\infty ,\\infty )[\/latex] into the smaller intervals [latex](\u2212\\infty ,0)[\/latex], [latex](0,1)[\/latex], [latex](1,2)[\/latex], and [latex](2,\\infty )[\/latex], and choose a test point from each interval to evaluate the sign of [latex]f^{\\prime}(x)[\/latex] in each of these smaller intervals. For example, let [latex]x=-1[\/latex], [latex]x=\\frac{1}{2}[\/latex], [latex]x=\\frac{3}{2}[\/latex], and [latex]x=3[\/latex] be the test points as shown in the following table.<\/p>\n<table id=\"fs-id1165042710566\" class=\"unnumbered\" summary=\"This table has four columns and five rows. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019(x) = (x2 \u2212 2x)\/(x \u2212 1)2 = x(x \u2212 2)\/(x \u2212 1)2, and Conclusion. Under the header row, the first column reads (\u2212\u221e, 0), (0, 1), (1, 2), and (2, \u221e). The second column reads x = \u22121, x = 1\/2, x = 3\/2, and x = 3. The third column reads (\u2212)(\u2212)\/+ = +, (+)(\u2212)\/+ = \u2212, (+)(\u2212)\/+ = \u2212, and (+)(+)\/+ = +. The fourth column reads f is increasing, f is decreasing, f is decreasing, and f is increasing.\">\n<thead>\n<tr valign=\"top\">\n<th>Interval<\/th>\n<th>Test Point<\/th>\n<th>Sign of [latex]f^{\\prime}(x)=\\frac{x^2-2x}{(x-1)^2}=\\frac{x(x-2)}{(x-1)^2}[\/latex]<\/th>\n<th>Conclusion<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex](\u2212\\infty ,0)[\/latex]<\/td>\n<td>[latex]x=-1[\/latex]<\/td>\n<td>[latex](\u2212)(\u2212)\/+=+[\/latex]<\/td>\n<td>[latex]f[\/latex] is increasing.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](0,1)[\/latex]<\/td>\n<td>[latex]x=1\/2[\/latex]<\/td>\n<td>[latex](+)(\u2212)\/+=\u2212[\/latex]<\/td>\n<td>[latex]f[\/latex] is decreasing.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](1,2)[\/latex]<\/td>\n<td>[latex]x=3\/2[\/latex]<\/td>\n<td>[latex](+)(\u2212)\/+=\u2212[\/latex]<\/td>\n<td>[latex]f[\/latex] is decreasing.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](2,\\infty )[\/latex]<\/td>\n<td>[latex]x=3[\/latex]<\/td>\n<td>[latex](+)(+)\/+=+[\/latex]<\/td>\n<td>[latex]f[\/latex] is increasing.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165042476453\">From this table, we see that [latex]f[\/latex] has a local maximum at [latex]x=0[\/latex] and a local minimum at [latex]x=2[\/latex]. The value of [latex]f[\/latex] at the local maximum is [latex]f(0)=0[\/latex] and the value of [latex]f[\/latex] at the local minimum is [latex]f(2)=4[\/latex]. Therefore, [latex](0,0)[\/latex] and [latex](2,4)[\/latex] are important points on the graph.<\/p>\n<p id=\"fs-id1165042464546\">Step 6. Calculate the second derivative:<\/p>\n<div id=\"fs-id1165042464549\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} f^{\\prime \\prime}(x) & =\\frac{(x-1)^2(2x-2)-(x^2-2x)(2(x-1))}{(x-1)^4} \\\\ & =\\frac{(x-1)[(x-1)(2x-2)-2(x^2-2x)]}{(x-1)^4} \\\\ & =\\frac{(x-1)(2x-2)-2(x^2-2x)}{(x-1)^3} \\\\ & =\\frac{2x^2-4x+2-(2x^2-4x)}{(x-1)^3} \\\\ & =\\frac{2}{(x-1)^3} \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165042652400\">We see that [latex]f^{\\prime \\prime}(x)[\/latex] is never zero or undefined for [latex]x[\/latex] in the domain of [latex]f[\/latex]. Since [latex]f[\/latex] is undefined at [latex]x=1[\/latex], to check concavity we just divide the interval [latex](\u2212\\infty ,\\infty )[\/latex] into the two smaller intervals [latex](\u2212\\infty ,1)[\/latex] and [latex](1,\\infty )[\/latex], and choose a test point from each interval to evaluate the sign of [latex]f^{\\prime \\prime}(x)[\/latex] in each of these intervals. The values [latex]x=0[\/latex] and [latex]x=2[\/latex] are possible test points as shown in the following table.<\/p>\n<table id=\"fs-id1165042652542\" class=\"unnumbered\" summary=\"This table has four columns and three rows. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019\u2019(x) = 2\/(x \u2212 1)3, and Conclusion. Under the header row, the first column reads (\u2212\u221e, 1) and (1, \u221e). The column row reads x = 0 and x = 2. The third column reads +\/\u2212 = \u2212 and +\/+ = +. The fourth column reads f is concave down and f is concave up.\">\n<thead>\n<tr valign=\"top\">\n<th>Interval<\/th>\n<th>Test Point<\/th>\n<th>Sign of [latex]f^{\\prime \\prime}(x)=\\frac{2}{(x-1)^3}[\/latex]<\/th>\n<th>Conclusion<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex](\u2212\\infty ,1)[\/latex]<\/td>\n<td>[latex]x=0[\/latex]<\/td>\n<td>[latex]+\/-=\u2212[\/latex]<\/td>\n<td>[latex]f[\/latex] is concave down.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](1,\\infty )[\/latex]<\/td>\n<td>[latex]x=2[\/latex]<\/td>\n<td>[latex]+\/+=+[\/latex]<\/td>\n<td>[latex]f[\/latex] is concave up.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165042606958\">From the information gathered, we arrive at the following graph for [latex]f.[\/latex]<\/p>\n<div style=\"width: 352px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211145\/CNX_Calc_Figure_04_06_017.jpg\" alt=\"The function f(x) = x2\/(x \u2212 1) is graphed. It has asymptotes y = x + 1 and x = 1.\" width=\"342\" height=\"346\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 27.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Sketching a Rational Function with an Oblique Asymptote.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/QhOY9VfIefo?controls=0&amp;start=552&amp;end=777&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266835\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266835\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.6LimitsAtInfinityAndAsymptotesPart2_552to777_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.6 Limits at Infinity and Asymptotes (part 2 &#8211; curve sketching)&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1165042606982\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find the oblique asymptote for [latex]f(x)=\\dfrac{3x^3-2x+1}{2x^2-4}[\/latex]<\/p>\n<div id=\"fs-id1165042606986\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q99083451\">Hint<\/span><\/p>\n<div id=\"q99083451\" class=\"hidden-answer\" style=\"display: none\"><\/div>\n<div>\n<p id=\"fs-id1165042607084\">Use long division of polynomials.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"exercise\">\n<p id=\"fs-id1165042607059\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042607059\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042607059\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y=\\frac{3}{2}x[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165042607090\" class=\"textbook exercises\">\n<h3>Example: Sketching the Graph of a Function with a Cusp<\/h3>\n<p>Sketch a graph of [latex]f(x)=(x-1)^{\\frac{2}{3}}[\/latex]<\/p>\n<div id=\"fs-id1165042607093\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042607145\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042607145\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042607145\">Step 1. Since the cube-root function is defined for all real numbers [latex]x[\/latex] and [latex](x-1)^{2\/3}=(\\sqrt[3]{x-1})^2[\/latex], the domain of [latex]f[\/latex] is all real numbers.<\/p>\n<p id=\"fs-id1165042607208\">Step 2: To find the [latex]y[\/latex]-intercept, evaluate [latex]f(0)[\/latex]. Since [latex]f(0)=1[\/latex], the [latex]y[\/latex]-intercept is [latex](0,1)[\/latex]. To find the [latex]x[\/latex]-intercept, solve [latex](x-1)^{2\/3}=0[\/latex]. The solution of this equation is [latex]x=1[\/latex], so the [latex]x[\/latex]-intercept is [latex](1,0)[\/latex].<\/p>\n<p id=\"fs-id1165042583208\">Step 3: Since [latex]\\underset{x\\to \\pm \\infty }{\\lim}(x-1)^{2\/3}=\\infty[\/latex], the function continues to grow without bound as [latex]x\\to \\infty[\/latex] and [latex]x\\to \u2212\\infty[\/latex].<\/p>\n<p id=\"fs-id1165042583285\">Step 4: The function has no vertical asymptotes.<\/p>\n<p id=\"fs-id1165042583288\">Step 5: To determine where [latex]f[\/latex] is increasing or decreasing, calculate [latex]f^{\\prime}[\/latex]. We find<\/p>\n<div id=\"fs-id1165042583307\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\frac{2}{3}(x-1)^{-1\/3}=\\frac{2}{3(x-1)^{1\/3}}[\/latex].<\/div>\n<p id=\"fs-id1165042583388\">This function is not zero anywhere, but it is undefined when [latex]x=1[\/latex]. Therefore, the only critical point is [latex]x=1[\/latex]. Divide the interval [latex](\u2212\\infty ,\\infty )[\/latex] into the smaller intervals [latex](\u2212\\infty ,1)[\/latex] and [latex](1,\\infty )[\/latex], and choose test points in each of these intervals to determine the sign of [latex]f^{\\prime}(x)[\/latex] in each of these smaller intervals. Let [latex]x=0[\/latex] and [latex]x=2[\/latex] be the test points as shown in the following table.<\/p>\n<table id=\"fs-id1165042504467\" class=\"unnumbered\" summary=\"This table has four columns and three rows. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019(x) = 2\/(3(x \u2212 1)1\/3), and Conclusion. Under the header row, the first column reads (\u2212\u221e, 1) and (1, \u221e). The second column reads x = 0 and x = 2. The third column reads +\/\u2212 = \u2212 and +\/+ = +. The fourth column reads f is decreasing and f is increasing.\">\n<thead>\n<tr valign=\"top\">\n<th>Interval<\/th>\n<th>Test Point<\/th>\n<th>Sign of [latex]f^{\\prime}(x)=\\frac{2}{3(x-1)^{1\/3}}[\/latex]<\/th>\n<th>Conclusion<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex](\u2212\\infty ,1)[\/latex]<\/td>\n<td>[latex]x=0[\/latex]<\/td>\n<td>[latex]+\/-=\u2212[\/latex]<\/td>\n<td>[latex]f[\/latex] is decreasing.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](1,\\infty )[\/latex]<\/td>\n<td>[latex]x=2[\/latex]<\/td>\n<td>[latex]+\/+=+[\/latex]<\/td>\n<td>[latex]f[\/latex] is increasing.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165042504734\">We conclude that [latex]f[\/latex] has a local minimum at [latex]x=1[\/latex]. Evaluating [latex]f[\/latex] at [latex]x=1[\/latex], we find that the value of [latex]f[\/latex] at the local minimum is zero. Note that [latex]f^{\\prime}(1)[\/latex] is undefined, so to determine the behavior of the function at this critical point, we need to examine [latex]\\underset{x\\to 1}{\\lim}f^{\\prime}(x)[\/latex]. Looking at the one-sided limits, we have<\/p>\n<div id=\"fs-id1165042510168\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 1^+}{\\lim}\\frac{2}{3(x-1)^{1\/3}}=\\infty[\/latex] and [latex]\\underset{x\\to 1^-}{\\lim}\\frac{2}{3(x-1)^{1\/3}}=\u2212\\infty[\/latex].<\/div>\n<p id=\"fs-id1165042510285\">Therefore, [latex]f[\/latex] has a cusp at [latex]x=1[\/latex].<\/p>\n<p id=\"fs-id1165042510304\">Step 6: To determine concavity, we calculate the second derivative of [latex]f[\/latex]:<\/p>\n<div id=\"fs-id1165042510313\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime \\prime}(x)=-\\frac{2}{9}(x-1)^{-4\/3}=\\frac{-2}{9(x-1)^{4\/3}}[\/latex].<\/div>\n<p id=\"fs-id1165042510398\">We find that [latex]f^{\\prime \\prime}(x)[\/latex] is defined for all [latex]x[\/latex], but is undefined when [latex]x=1[\/latex]. Therefore, divide the interval [latex](\u2212\\infty ,\\infty )[\/latex] into the smaller intervals [latex](\u2212\\infty ,1)[\/latex] and [latex](1,\\infty )[\/latex], and choose test points to evaluate the sign of [latex]f^{\\prime \\prime}(x)[\/latex] in each of these intervals. As we did earlier, let [latex]x=0[\/latex] and [latex]x=2[\/latex] be test points as shown in the following table.<\/p>\n<table id=\"fs-id1165042510530\" class=\"unnumbered\" summary=\"This table has four columns and three rows. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019\u2019(x) = \u22122\/(9(x \u2212 1)4\/3), and Conclusion. Under the header row, the first column reads (\u2212\u221e, 1) and (1, \u221e). The second column reads x = 0 and x = 2. The third column reads \u2212\/+ = \u2212 and \u2212\/+ = \u2212. The fourth column reads f is concave down and f is concave down.\">\n<thead>\n<tr valign=\"top\">\n<th>Interval<\/th>\n<th>Test Point<\/th>\n<th>Sign of [latex]f^{\\prime \\prime}(x)=\\frac{-2}{9(x-1)^{4\/3}}[\/latex]<\/th>\n<th>Conclusion<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex](\u2212\\infty ,1)[\/latex]<\/td>\n<td>[latex]x=0[\/latex]<\/td>\n<td>[latex]\u2212\/+=\u2212[\/latex]<\/td>\n<td>[latex]f[\/latex] is concave down.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](1,\\infty )[\/latex]<\/td>\n<td>[latex]x=2[\/latex]<\/td>\n<td>[latex]\u2212\/+=\u2212[\/latex]<\/td>\n<td>[latex]f[\/latex] is concave down.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165042643969\">From this table, we conclude that [latex]f[\/latex] is concave down everywhere. Combining all of this information, we arrive at the following graph for [latex]f[\/latex].<\/p>\n<div style=\"width: 614px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211148\/CNX_Calc_Figure_04_06_018.jpg\" alt=\"The function f(x) = (x \u2212 1)2\/3 is graphed. It touches the x axis at x = 1, where it comes to something of a sharp point and then flairs out on either side.\" width=\"604\" height=\"272\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 28.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Sketching the Graph of a Function with a Cusp.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/QhOY9VfIefo?controls=0&amp;start=781&amp;end=962&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.6LimitsAtInfinityAndAsymptotesPart2_781to962_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.6 Limits at Infinity and Asymptotes (part 2 &#8211; curve sketching)&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Consider the function [latex]f(x)=5-x^{\\frac{2}{3}}[\/latex]. Determine the point on the graph where a cusp is located. Determine the end behavior of [latex]f[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q44668822\">Hint<\/span><\/p>\n<div id=\"q44668822\" class=\"hidden-answer\" style=\"display: none\">\n<p>A function [latex]f[\/latex] has a cusp at a point [latex]a[\/latex] if [latex]f(a)[\/latex] exists, [latex]f^{\\prime}(a)[\/latex] is undefined, one of the one-sided limits as [latex]x\\to a[\/latex] of [latex]f^{\\prime}(x)[\/latex] is [latex]+\\infty[\/latex], and the other one-sided limit is [latex]\u2212\\infty[\/latex].<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1165042644060\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1165042644060\" class=\"hidden-answer\" style=\"display: none\">\n<p>The function [latex]f[\/latex] has a cusp at [latex](0,5)[\/latex]: [latex]\\underset{x\\to 0^-}{\\lim}f^{\\prime}(x)=\\infty[\/latex], [latex]\\underset{x\\to 0^+}{\\lim}f^{\\prime}(x)=\u2212\\infty[\/latex]. For end behavior, [latex]\\underset{x\\to \\pm \\infty }{\\lim}f(x)=\u2212\\infty[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-409\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>4.6 Limits at Infinity and Asymptotes (part 2 - curve sketching). <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 1. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":22,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"4.6 Limits at Infinity and Asymptotes (part 2 - curve sketching)\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-409","chapter","type-chapter","status-publish","hentry"],"part":48,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/409","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":19,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/409\/revisions"}],"predecessor-version":[{"id":4916,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/409\/revisions\/4916"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/48"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/409\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=409"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=409"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=409"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=409"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}