Step 1. Since [latex]f[/latex] is a polynomial, the domain is the set of all real numbers.

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]

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 −\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]. 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

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].

Step 4. Since [latex]f[/latex] is a polynomial function, it does not have any vertical asymptotes.

Step 5. The first derivative of [latex]f[/latex] is

Therefore, [latex]f[/latex] has two critical points: [latex]x=1,-1[/latex]. Divide the interval [latex](−\infty ,\infty)[/latex] into the three smaller intervals: [latex](−\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.

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].

Step 6. The second derivative of [latex]f[/latex] is

The second derivative is zero at [latex]x=0[/latex]. Therefore, to determine the concavity of [latex]f[/latex], divide the interval [latex](−\infty ,\infty)[/latex] into the smaller intervals [latex](−\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.

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—namely, [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—combined 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].

Combining this information, we arrive at the graph of [latex]f(x)=(x-1)^2 (x+2)[/latex] shown in the following graph.

Figure 23.

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].

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.

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

Therefore, [latex]f[/latex] has a horizontal asymptote of [latex]y=-1[/latex] as [latex]x\to \infty [/latex] and [latex]x\to −\infty[/latex].

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 −1}{\lim}f(x)[/latex]. By looking at each one-sided limit as [latex]x\to 1[/latex], we see that

In addition, by looking at each one-sided limit as [latex]x\to −1[/latex], we find that

Step 5. Calculate the first derivative:

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](−\infty ,\infty )[/latex] into four smaller intervals: [latex](−\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.

From this analysis, we conclude that [latex]f[/latex] has a local minimum at [latex]x=0[/latex] but no local maximum.

Step 6. Calculate the second derivative:

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](−\infty ,\infty )[/latex] into the three smaller intervals [latex](−\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.

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].

Figure 25.

Step 1. The domain of [latex]f[/latex] is the set of all real numbers [latex]x[/latex] except [latex]x=1[/latex].

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.

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

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].

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

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.

Step 5. Calculate the first derivative:

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](−\infty ,\infty )[/latex] into the smaller intervals [latex](−\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.

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.

Step 6. Calculate the second derivative:

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](−\infty ,\infty )[/latex] into the two smaller intervals [latex](−\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.

From the information gathered, we arrive at the following graph for [latex]f.[/latex]

Figure 27.

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.

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].

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 −\infty[/latex].

Step 4: The function has no vertical asymptotes.

Step 5: To determine where [latex]f[/latex] is increasing or decreasing, calculate [latex]f^{\prime}[/latex]. We find

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](−\infty ,\infty )[/latex] into the smaller intervals [latex](−\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.

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

Therefore, [latex]f[/latex] has a cusp at [latex]x=1[/latex].

Step 6: To determine concavity, we calculate the second derivative of [latex]f[/latex]:

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](−\infty ,\infty )[/latex] into the smaller intervals [latex](−\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.

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].

Figure 28.