{"id":458,"date":"2021-02-04T15:28:04","date_gmt":"2021-02-04T15:28:04","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&p=458"},"modified":"2021-04-08T02:52:26","modified_gmt":"2021-04-08T02:52:26","slug":"problem-set-continuity","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/problem-set-continuity\/","title":{"raw":"Problem Set: Continuity","rendered":"Problem Set: Continuity"},"content":{"raw":"

For the following exercises (1-8), determine the point(s), if any, at which each function is discontinuous. Classify any discontinuity as jump, removable, infinite, or other.<\/p>\r\n\r\n

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1.\u00a0<\/strong>[latex]f(x)=\\dfrac{1}{\\sqrt{x}}[\/latex]<\/p>\r\n[reveal-answer q=\"955865\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"955865\"]\r\nThe function is defined for all [latex]x[\/latex] in the interval [latex](0,\\infty)[\/latex].\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

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2.\u00a0<\/strong>[latex]f(x)=\\dfrac{2}{x^2+1}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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3.\u00a0<\/strong>[latex]f(x)=\\dfrac{x}{x^2-x}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170573590405\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573590405\"]\r\n

Removable discontinuity at [latex]x=0[\/latex]; infinite discontinuity at [latex]x=1[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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4.\u00a0<\/strong>[latex]g(t)=t^{-1}+1[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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5.\u00a0<\/strong>[latex]f(x)=\\dfrac{5}{e^x-2}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170573586367\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573586367\"]\r\n

Infinite discontinuity at [latex]x=\\ln 2[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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6.\u00a0<\/strong>[latex]f(x)=\\dfrac{|x-2|}{x-2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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7.\u00a0<\/strong>[latex]H(x)= \\tan 2x[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170571047553\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571047553\"]\r\n

Infinite discontinuities at [latex]x=\\frac{(2k+1)\\pi}{4}[\/latex], for [latex]k=0, \\, \\pm 1, \\, \\pm 2, \\, \\pm 3, \\cdots[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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8.\u00a0<\/strong>[latex]f(t)=\\dfrac{t+3}{t^2+5t+6}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

For the following exercises (9-14), decide if the function continuous at the given point. If it is discontinuous, what type of discontinuity is it?<\/p>\r\n\r\n

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9.\u00a0<\/strong>[latex]f(x)=\\dfrac{2x^2-5x+3}{x-1}[\/latex] at [latex]x=1[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170573331459\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573331459\"]\r\n

No. It is a removable discontinuity.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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10.\u00a0<\/strong>[latex]h(\\theta)=\\dfrac{\\sin \\theta - \\cos \\theta}{\\tan \\theta}[\/latex] at [latex]\\theta =\\pi[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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11.\u00a0<\/strong>[latex]g(u)=\\begin{cases} \\dfrac{6u^2+u-2}{2u-1} & \\text{ if } \\, u \\ne \\frac{1}{2} \\\\ \\dfrac{7}{2} & \\text{ if } \\, u = \\frac{1}{2} \\end{cases}[\/latex] at [latex]u=\\frac{1}{2}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170571120270\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571120270\"]\r\n

Yes. It is continuous.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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12.\u00a0<\/strong>[latex]f(y)=\\dfrac{\\sin(\\pi y)}{\\tan(\\pi y)}[\/latex], at [latex]y=1[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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13.\u00a0<\/strong>[latex]f(x)=\\begin{cases} x^2-e^x & \\text{ if } \\, x < 0 \\\\ x-1 & \\text{ if } \\, x \\ge 0 \\end{cases}[\/latex] at [latex]x=0[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170573381211\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573381211\"]\r\n

Yes. It is continuous.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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14.\u00a0<\/strong>[latex]f(x)=\\begin{cases} x \\sin x & \\text{ if } \\, x \\le \\pi \\\\ x \\tan x & \\text{ if } \\, x > \\pi \\end{cases}[\/latex] at [latex]x=\\pi[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

In the following exercises (15-19), find the value(s) of [latex]k[\/latex] that makes each function continuous over the given interval.<\/p>\r\n\r\n

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15.\u00a0<\/strong>[latex]f(x)=\\begin{cases} 3x+2 & \\text{ if } \\, x < k \\\\ 2x-3 & \\text{ if } \\, k \\le x \\le 8 \\end{cases}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170573440208\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573440208\"]\r\n

[latex]k=-5[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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16.\u00a0<\/strong>[latex]f(\\theta)=\\begin{cases} \\sin \\theta & \\text{ if } \\, 0 \\le \\theta < \\frac{\\pi}{2} \\\\ \\cos (\\theta + k) & \\text{ if } \\, \\frac{\\pi}{2} \\le \\theta \\le \\pi \\end{cases}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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17.\u00a0<\/strong>[latex]f(x)=\\begin{cases} \\dfrac{x^2+3x+2}{x+2} & \\text{ if } \\, x \\ne -2 \\\\ k & \\text{ if } \\, x = -2 \\end{cases}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170573502717\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573502717\"]\r\n

[latex]k=-1[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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18.\u00a0<\/strong>[latex]f(x)=\\begin{cases} e^{kx} & \\text{ if } \\, 0 \\le x < 4 \\\\ x+3 & \\text{ if } \\, 4 \\le x \\le 8 \\end{cases}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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19.\u00a0<\/strong>[latex]f(x)=\\begin{cases} \\sqrt{kx} & \\text{ if } \\, 0 \\le x \\le 3 \\\\ x+1 & \\text{ if } \\, 3 < x \\le 10 \\end{cases}[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170571050054\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571050054\"]\r\n

[latex]k=\\frac{16}{3}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

In the following exercises (20-21), use the Intermediate Value Theorem (IVT).<\/p>\r\n\r\n

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20.\u00a0<\/strong>Let [latex]h(x)=\\begin{cases} 3x^2-4 & \\text{ if } \\, x \\le 2 \\\\ 5+4x & \\text{ if } \\, x > 2 \\end{cases}[\/latex] Over the interval [latex][0,4][\/latex], there is no value of [latex]x[\/latex] such that [latex]h(x)=10[\/latex], although [latex]h(0)<10[\/latex] and [latex]h(4)>10[\/latex]. Explain why this does not contradict the IVT.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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21.\u00a0<\/strong>A particle moving along a line has at each time [latex]t[\/latex] a position function [latex]s(t)[\/latex], which is continuous. Assume [latex]s(2)=5[\/latex] and [latex]s(5)=2[\/latex]. Another particle moves such that its position is given by [latex]h(t)=s(t)-t[\/latex]. Explain why there must be a value [latex]c[\/latex] for [latex]2<c<5[\/latex] such that [latex]h(c)=0[\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1170570998991\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170570998991\"]\r\n

Since both [latex]s[\/latex] and [latex]y=t[\/latex] are continuous everywhere, then [latex]h(t)=s(t)-t[\/latex] is continuous everywhere and, in particular, it is continuous over the closed interval [latex][2,5][\/latex]. Also, [latex]h(2)=3>0[\/latex] and [latex]h(5)=-3<0[\/latex]. Therefore, by the IVT, there is a value [latex]x=c[\/latex] such that [latex]h(c)=0[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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22. [T]<\/strong> Use the statement \u201cThe cosine of [latex]t[\/latex] is equal to [latex]t[\/latex] cubed.\u201d<\/p>\r\n\r\n

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  1. Write the statement as a mathematical equation.<\/li>\r\n \t
  2. Prove that the equation in part (a) has at least one real solution.<\/li>\r\n \t
  3. Use a calculator to find an interval of length 0.01 that contains a solution of the equation.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n
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    23.\u00a0<\/strong>Apply the IVT to determine whether [latex]2^x=x^3[\/latex] has a solution in one of the intervals [latex][1.25,1.375][\/latex] or [latex][1.375,1.5][\/latex]. Briefly explain your response for each interval.<\/p>\r\n[reveal-answer q=\"fs-id1170570998210\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170570998210\"]\r\n

    The function [latex]f(x)=2^x-x^3[\/latex] is continuous over the interval [latex][1.25,1.375][\/latex] and has opposite signs at the endpoints.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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    24.\u00a0<\/strong>Consider the graph of the function [latex]y=f(x)[\/latex] shown in the following graph.<\/p>\r\n\"A<\/span>\r\n

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    1. Find all values for which the function is discontinuous.<\/li>\r\n \t
    2. For each value in part (a), use the formal definition of continuity to explain why the function is discontinuous at that value.<\/li>\r\n \t
    3. Classify each discontinuity as either jump, removable, or infinite.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n
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      25.\u00a0<\/strong>Let [latex]f(x)=\\begin{cases} 3x & \\text{ if } \\, x > 1 \\\\ x^3 & \\text{ if } \\, x < 1 \\end{cases}[\/latex]<\/p>\r\n\r\n

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      1. Sketch the graph of [latex]f[\/latex].<\/li>\r\n \t
      2. Is it possible to find a value [latex]k[\/latex] such that [latex]f(1)=k[\/latex], which makes [latex]f(x)[\/latex] continuous for all real numbers? Briefly explain.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1170571100273\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571100273\"]\r\n

        a.<\/p>\r\n\"A<\/span>\r\nb. It is not possible to redefine [latex]f(1)[\/latex] since the discontinuity is a jump discontinuity.[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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        26.\u00a0<\/strong>Let [latex]f(x)=\\dfrac{x^4-1}{x^2-1}[\/latex] for [latex]x\\ne -1,1[\/latex].<\/p>\r\n\r\n

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        1. Sketch the graph of [latex]f[\/latex].<\/li>\r\n \t
        2. Is it possible to find values [latex]k_1[\/latex] and [latex]k_2[\/latex] such that [latex]f(-1)=k_1[\/latex] and [latex]f(1)=k_2[\/latex], and that makes [latex]f(x)[\/latex] continuous for all real numbers? Briefly explain.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n
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          27.\u00a0<\/strong>Sketch the graph of the function [latex]y=f(x)[\/latex] with properties 1 through 7.<\/p>\r\n\r\n

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          1. The domain of [latex]f[\/latex] is [latex](\u2212\\infty,+\\infty)[\/latex].<\/li>\r\n \t
          2. [latex]f[\/latex] has an infinite discontinuity at [latex]x=-6[\/latex].<\/li>\r\n \t
          3. [latex]f(-6)=3[\/latex]<\/li>\r\n \t
          4. [latex]\\underset{x\\to -3^-}{\\lim}f(x)=\\underset{x\\to -3^+}{\\lim}f(x)=2[\/latex]<\/li>\r\n \t
          5. [latex]f(-3)=3[\/latex]<\/li>\r\n \t
          6. [latex]f[\/latex] is left continuous but not right continuous at [latex]x=3[\/latex].<\/li>\r\n \t
          7. [latex]\\underset{x\\to -\\infty}{\\lim}f(x)=\u2212\\infty[\/latex] and [latex]\\underset{x\\to +\\infty}{\\lim}f(x)=+\\infty[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1170570982616\"]Show Solution[\/reveal-answer]\r\n
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            28.\u00a0<\/strong>Sketch the graph of the function [latex]y=f(x)[\/latex] with properties 1 through 4.<\/p>\r\n\r\n

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            1. The domain of [latex]f[\/latex] is [latex][0,5][\/latex].<\/li>\r\n \t
            2. [latex]\\underset{x\\to 1^+}{\\lim}f(x)[\/latex] and [latex]\\underset{x\\to 1^-}{\\lim}f(x)[\/latex] exist and are equal.<\/li>\r\n \t
            3. [latex]f(x)[\/latex] is left continuous but not continuous at [latex]x=2[\/latex], and right continuous but not continuous at [latex]x=3[\/latex].<\/li>\r\n \t
            4. [latex]f(x)[\/latex] has a removable discontinuity at [latex]x=1[\/latex], a jump discontinuity at [latex]x=2[\/latex], and the following limits hold: [latex]\\underset{x\\to 3^-}{\\lim}f(x)=\u2212\\infty[\/latex] and [latex]\\underset{x\\to 3^+}{\\lim}f(x)=2[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n

              In the following exercises (29-30), suppose [latex]y=f(x)[\/latex] is defined for all [latex]x[\/latex]. For each description, sketch a graph with the indicated property.<\/p>\r\n\r\n

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              29.\u00a0<\/strong>Discontinuous at [latex]x=1[\/latex] with [latex]\\underset{x\\to -1}{\\lim}f(x)=-1[\/latex] and [latex]\\underset{x\\to 2}{\\lim}f(x)=4[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170571123483\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571123483\"]\r\n

              Answers may vary; see the following example:<\/p>\r\n\"The<\/span>[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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              30.\u00a0<\/strong>Discontinuous at [latex]x=2[\/latex] but continuous elsewhere with [latex]\\underset{x\\to 0}{\\lim}f(x)=\\frac{1}{2}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

              Determine whether each of the given statements is true (31-37). Justify your responses with an explanation or counterexample.<\/p>\r\n\r\n

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              31.\u00a0<\/strong>[latex]f(t)=\\dfrac{2}{e^t-e^{-t}}[\/latex] is continuous everywhere.<\/p>\r\n[reveal-answer q=\"fs-id1170573404350\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573404350\"]\r\n

              False. It is continuous over [latex](\u2212\\infty,0) \\cup (0,\\infty)[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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              32.\u00a0<\/strong>If the left- and right-hand limits of [latex]f(x)[\/latex] as [latex]x\\to a[\/latex] exist and are equal, then [latex]f[\/latex] cannot be discontinuous at [latex]x=a[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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              33.\u00a0<\/strong>If a function is not continuous at a point, then it is not defined at that point.<\/p>\r\n[reveal-answer q=\"fs-id1170571287020\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571287020\"]\r\n

              False. Consider [latex]f(x)=\\begin{cases} x & \\text{ if } \\, x \\ne 0 \\\\ 4 & \\text{ if } \\, x = 0 \\end{cases}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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              34.\u00a0<\/strong>According to the IVT, [latex]\\cos x - \\sin x - x = 2[\/latex] has a solution over the interval [latex][-1,1][\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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              35.\u00a0<\/strong>If [latex]f(x)[\/latex] is continuous such that [latex]f(a)[\/latex] and [latex]f(b)[\/latex] have opposite signs, then [latex]f(x)=0[\/latex] has exactly one solution in [latex][a,b][\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1170573519289\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573519289\"]\r\n

              False. Consider [latex]f(x)= \\cos (x)[\/latex] on [latex][-\\pi, 2\\pi][\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

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              36.\u00a0<\/strong>The function [latex]f(x)=\\dfrac{x^2-4x+3}{x^2-1}[\/latex] is continuous over the interval [latex][0,3][\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

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              37.\u00a0<\/strong>If [latex]f(x)[\/latex] is continuous everywhere and [latex]f(a), f(b)>0[\/latex], then there is no root of [latex]f(x)[\/latex] in the interval [latex][a,b][\/latex].<\/p>\r\n[reveal-answer q=\"fs-id1170571197414\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571197414\"]\r\n

              False. The IVT does not<\/em> work in reverse! Consider [latex](x-1)^2[\/latex] over the interval [latex][-2,2][\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

              The following problems (38-39) consider the scalar form of Coulomb\u2019s law, which describes the electrostatic force between two point charges, such as electrons. It is given by the equation [latex]F(r)=k_e\\dfrac{|q_1q_2|}{r^2}[\/latex], where [latex]k_e[\/latex] is Coulomb\u2019s constant, [latex]q_i[\/latex] are the magnitudes of the charges of the two particles, and [latex]r[\/latex] is the distance between the two particles.<\/p>\r\n\r\n

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              38. [T]<\/strong> To simplify the calculation of a model with many interacting particles, after some threshold value [latex]r=R[\/latex], we approximate [latex]F[\/latex] as zero.<\/p>\r\n\r\n

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              1. Explain the physical reasoning behind this assumption.<\/li>\r\n \t
              2. What is the force equation?<\/li>\r\n \t
              3. Evaluate the force [latex]F[\/latex] using both Coulomb\u2019s law and our approximation, assuming two protons with a charge magnitude of [latex]1.6022 \\times 10^{-19} \\, \\text{coulombs (C)}[\/latex], and the Coulomb constant [latex]k_e = 8.988 \\times 10^9 \\, \\text{Nm}^2\/\\text{C}^2[\/latex] are 1 m apart. Also, assume [latex]R<1\\text{m}[\/latex]. How much inaccuracy does our approximation generate? Is our approximation reasonable?<\/li>\r\n \t
              4. Is there any finite value of [latex]R[\/latex] for which this system remains continuous at [latex]R[\/latex]?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<\/div>\r\n
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                39. [T]\u00a0<\/strong>Instead of making the force 0 at [latex]R[\/latex], instead we let the force be [latex]10^{-20}[\/latex]\u00a0for [latex]r\\ge R[\/latex]. Assume two protons, which have a magnitude of charge [latex]1.6022 \\times 10^{-19} \\, \\text{C}[\/latex], and the Coulomb constant [latex]k_e=8.988 \\times 10^9 \\, \\text{Nm}^2\/\\text{C}^2[\/latex]. Is there a value [latex]R[\/latex] that can make this system continuous? If so, find it.<\/p>\r\n[reveal-answer q=\"fs-id1170573420081\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573420081\"]\r\n

                [latex]R=0.0001519 \\, \\text{m}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

                Recall the discussion on spacecraft from the Why It Matters. The following problems (40-42) consider a rocket launch from Earth\u2019s surface. The force of gravity on the rocket is given by [latex]F(d)=\\frac{-mk}{d^2}[\/latex], where [latex]m[\/latex] is the mass of the rocket, [latex]d[\/latex] is the distance of the rocket from the center of Earth, and [latex]k[\/latex] is a constant.<\/p>\r\n\r\n

                \r\n
                \r\n

                40. [T]<\/strong> Determine the value and units of [latex]k[\/latex] given that the mass of the rocket on Earth is 3 million kg. (Hint<\/em>: The distance from the center of Earth to its surface is 6378 km.)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

                \r\n
                \r\n

                41. [T]<\/strong> After a certain distance [latex]D[\/latex] has passed, the gravitational effect of Earth becomes quite negligible, so we can approximate the force function by [latex]F(d)=\\begin{cases} -\\dfrac{mk}{d^2} & \\text{ if } \\, d < D \\\\ 10,000 & \\text{ if } \\, d \\ge D \\end{cases}[\/latex] Find the necessary condition [latex]D[\/latex] such that the force function remains continuous.<\/p>\r\n[reveal-answer q=\"fs-id1170570974552\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170570974552\"]\r\n

                [latex]D=63.78[\/latex] km<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

                \r\n
                \r\n

                42.\u00a0<\/strong>As the rocket travels away from Earth\u2019s surface, there is a distance [latex]D[\/latex] where the rocket sheds some of its mass, since it no longer needs the excess fuel storage. We can write this function as [latex]F(d)=\\begin{cases} -\\dfrac{m_1 k}{d^2} & \\text{ if } \\, d < D \\\\ -\\dfrac{m_2 k}{d^2} & \\text{ if } \\, d \\ge D \\end{cases}[\/latex] Is there a [latex]D[\/latex]\u00a0value such that this function is continuous, assuming [latex]m_1 \\ne m_2[\/latex]?<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

                Prove the following functions are continuous everywhere (43-45).<\/p>\r\n\r\n

                \r\n
                \r\n

                43.\u00a0<\/strong>[latex]f(\\theta) = \\sin \\theta[\/latex]<\/p>\r\n[reveal-answer q=\"fs-id1170571276973\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571276973\"]\r\n

                For all values of [latex]a, \\, f(a)[\/latex] is defined, [latex]\\underset{\\theta \\to a}{\\lim}f(\\theta)[\/latex] exists, and [latex]\\underset{\\theta \\to a}{\\lim}f(\\theta)=f(a)[\/latex]. Therefore, [latex]f(\\theta)[\/latex] is continuous everywhere.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n

                \r\n
                \r\n

                44.\u00a0<\/strong>[latex]g(x)=|x|[\/latex]<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

                \r\n
                \r\n

                45.\u00a0<\/strong>Where is [latex]f(x)=\\begin{cases} 0 & \\text{ if } \\, x \\, \\text{is irrational} \\\\ 1 & \\text{ if } \\, x \\, \\text{is rational} \\end{cases}[\/latex] continuous?<\/p>\r\n[reveal-answer q=\"fs-id1170571258410\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170571258410\"]\r\n

                Nowhere<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>","rendered":"

                For the following exercises (1-8), determine the point(s), if any, at which each function is discontinuous. Classify any discontinuity as jump, removable, infinite, or other.<\/p>\n

                \n

                1.\u00a0<\/strong>[latex]f(x)=\\dfrac{1}{\\sqrt{x}}[\/latex]<\/p>\n

                Show Solution<\/span><\/p>\n
                \nThe function is defined for all [latex]x[\/latex] in the interval [latex](0,\\infty)[\/latex].\n<\/div>\n<\/div>\n<\/div>\n
                \n
                \n

                2.\u00a0<\/strong>[latex]f(x)=\\dfrac{2}{x^2+1}[\/latex]<\/p>\n<\/div>\n<\/div>\n

                \n
                \n

                3.\u00a0<\/strong>[latex]f(x)=\\dfrac{x}{x^2-x}[\/latex]<\/p>\n

                Show Solution<\/span><\/p>\n
                \n

                Removable discontinuity at [latex]x=0[\/latex]; infinite discontinuity at [latex]x=1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

                \n
                \n

                4.\u00a0<\/strong>[latex]g(t)=t^{-1}+1[\/latex]<\/p>\n<\/div>\n<\/div>\n

                \n
                \n

                5.\u00a0<\/strong>[latex]f(x)=\\dfrac{5}{e^x-2}[\/latex]<\/p>\n

                Show Solution<\/span><\/p>\n
                \n

                Infinite discontinuity at [latex]x=\\ln 2[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

                \n
                \n

                6.\u00a0<\/strong>[latex]f(x)=\\dfrac{|x-2|}{x-2}[\/latex]<\/p>\n<\/div>\n<\/div>\n

                \n
                \n

                7.\u00a0<\/strong>[latex]H(x)= \\tan 2x[\/latex]<\/p>\n

                Show Solution<\/span><\/p>\n
                \n

                Infinite discontinuities at [latex]x=\\frac{(2k+1)\\pi}{4}[\/latex], for [latex]k=0, \\, \\pm 1, \\, \\pm 2, \\, \\pm 3, \\cdots[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

                \n
                \n

                8.\u00a0<\/strong>[latex]f(t)=\\dfrac{t+3}{t^2+5t+6}[\/latex]<\/p>\n<\/div>\n<\/div>\n

                For the following exercises (9-14), decide if the function continuous at the given point. If it is discontinuous, what type of discontinuity is it?<\/p>\n

                \n
                \n

                9.\u00a0<\/strong>[latex]f(x)=\\dfrac{2x^2-5x+3}{x-1}[\/latex] at [latex]x=1[\/latex]<\/p>\n

                Show Solution<\/span><\/p>\n
                \n

                No. It is a removable discontinuity.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

                \n
                \n

                10.\u00a0<\/strong>[latex]h(\\theta)=\\dfrac{\\sin \\theta - \\cos \\theta}{\\tan \\theta}[\/latex] at [latex]\\theta =\\pi[\/latex]<\/p>\n<\/div>\n<\/div>\n

                \n
                \n

                11.\u00a0<\/strong>[latex]g(u)=\\begin{cases} \\dfrac{6u^2+u-2}{2u-1} & \\text{ if } \\, u \\ne \\frac{1}{2} \\\\ \\dfrac{7}{2} & \\text{ if } \\, u = \\frac{1}{2} \\end{cases}[\/latex] at [latex]u=\\frac{1}{2}[\/latex]<\/p>\n

                Show Solution<\/span><\/p>\n
                \n

                Yes. It is continuous.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

                \n
                \n

                12.\u00a0<\/strong>[latex]f(y)=\\dfrac{\\sin(\\pi y)}{\\tan(\\pi y)}[\/latex], at [latex]y=1[\/latex]<\/p>\n<\/div>\n<\/div>\n

                \n
                \n

                13.\u00a0<\/strong>[latex]f(x)=\\begin{cases} x^2-e^x & \\text{ if } \\, x < 0 \\\\ x-1 & \\text{ if } \\, x \\ge 0 \\end{cases}[\/latex] at [latex]x=0[\/latex]<\/p>\n

                Show Solution<\/span><\/p>\n
                \n

                Yes. It is continuous.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

                \n
                \n

                14.\u00a0<\/strong>[latex]f(x)=\\begin{cases} x \\sin x & \\text{ if } \\, x \\le \\pi \\\\ x \\tan x & \\text{ if } \\, x > \\pi \\end{cases}[\/latex] at [latex]x=\\pi[\/latex]<\/p>\n<\/div>\n<\/div>\n

                In the following exercises (15-19), find the value(s) of [latex]k[\/latex] that makes each function continuous over the given interval.<\/p>\n

                \n
                \n

                15.\u00a0<\/strong>[latex]f(x)=\\begin{cases} 3x+2 & \\text{ if } \\, x < k \\\\ 2x-3 & \\text{ if } \\, k \\le x \\le 8 \\end{cases}[\/latex]<\/p>\n

                Show Solution<\/span><\/p>\n
                \n

                [latex]k=-5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

                \n
                \n

                16.\u00a0<\/strong>[latex]f(\\theta)=\\begin{cases} \\sin \\theta & \\text{ if } \\, 0 \\le \\theta < \\frac{\\pi}{2} \\\\ \\cos (\\theta + k) & \\text{ if } \\, \\frac{\\pi}{2} \\le \\theta \\le \\pi \\end{cases}[\/latex]<\/p>\n<\/div>\n<\/div>\n

                \n
                \n

                17.\u00a0<\/strong>[latex]f(x)=\\begin{cases} \\dfrac{x^2+3x+2}{x+2} & \\text{ if } \\, x \\ne -2 \\\\ k & \\text{ if } \\, x = -2 \\end{cases}[\/latex]<\/p>\n

                Show Solution<\/span><\/p>\n
                \n

                [latex]k=-1[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

                \n
                \n

                18.\u00a0<\/strong>[latex]f(x)=\\begin{cases} e^{kx} & \\text{ if } \\, 0 \\le x < 4 \\\\ x+3 & \\text{ if } \\, 4 \\le x \\le 8 \\end{cases}[\/latex]<\/p>\n<\/div>\n<\/div>\n

                \n
                \n

                19.\u00a0<\/strong>[latex]f(x)=\\begin{cases} \\sqrt{kx} & \\text{ if } \\, 0 \\le x \\le 3 \\\\ x+1 & \\text{ if } \\, 3 < x \\le 10 \\end{cases}[\/latex]<\/p>\n

                Show Solution<\/span><\/p>\n
                \n

                [latex]k=\\frac{16}{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

                In the following exercises (20-21), use the Intermediate Value Theorem (IVT).<\/p>\n

                \n
                \n

                20.\u00a0<\/strong>Let [latex]h(x)=\\begin{cases} 3x^2-4 & \\text{ if } \\, x \\le 2 \\\\ 5+4x & \\text{ if } \\, x > 2 \\end{cases}[\/latex] Over the interval [latex][0,4][\/latex], there is no value of [latex]x[\/latex] such that [latex]h(x)=10[\/latex], although [latex]h(0)<10[\/latex] and [latex]h(4)>10[\/latex]. Explain why this does not contradict the IVT.<\/p>\n<\/div>\n<\/div>\n

                \n
                \n

                21.\u00a0<\/strong>A particle moving along a line has at each time [latex]t[\/latex] a position function [latex]s(t)[\/latex], which is continuous. Assume [latex]s(2)=5[\/latex] and [latex]s(5)=2[\/latex]. Another particle moves such that its position is given by [latex]h(t)=s(t)-t[\/latex]. Explain why there must be a value [latex]c[\/latex] for [latex]2\n

                Show Solution<\/span><\/p>\n
                \n

                Since both [latex]s[\/latex] and [latex]y=t[\/latex] are continuous everywhere, then [latex]h(t)=s(t)-t[\/latex] is continuous everywhere and, in particular, it is continuous over the closed interval [latex][2,5][\/latex]. Also, [latex]h(2)=3>0[\/latex] and [latex]h(5)=-3<0[\/latex]. Therefore, by the IVT, there is a value [latex]x=c[\/latex] such that [latex]h(c)=0[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

                \n
                \n

                22. [T]<\/strong> Use the statement \u201cThe cosine of [latex]t[\/latex] is equal to [latex]t[\/latex] cubed.\u201d<\/p>\n

                  \n
                1. Write the statement as a mathematical equation.<\/li>\n
                2. Prove that the equation in part (a) has at least one real solution.<\/li>\n
                3. Use a calculator to find an interval of length 0.01 that contains a solution of the equation.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n
                  \n
                  \n

                  23.\u00a0<\/strong>Apply the IVT to determine whether [latex]2^x=x^3[\/latex] has a solution in one of the intervals [latex][1.25,1.375][\/latex] or [latex][1.375,1.5][\/latex]. Briefly explain your response for each interval.<\/p>\n

                  Show Solution<\/span><\/p>\n
                  \n

                  The function [latex]f(x)=2^x-x^3[\/latex] is continuous over the interval [latex][1.25,1.375][\/latex] and has opposite signs at the endpoints.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

                  \n
                  \n

                  24.\u00a0<\/strong>Consider the graph of the function [latex]y=f(x)[\/latex] shown in the following graph.<\/p>\n

                  \"A<\/span><\/p>\n

                    \n
                  1. Find all values for which the function is discontinuous.<\/li>\n
                  2. For each value in part (a), use the formal definition of continuity to explain why the function is discontinuous at that value.<\/li>\n
                  3. Classify each discontinuity as either jump, removable, or infinite.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n
                    \n
                    \n

                    25.\u00a0<\/strong>Let [latex]f(x)=\\begin{cases} 3x & \\text{ if } \\, x > 1 \\\\ x^3 & \\text{ if } \\, x < 1 \\end{cases}[\/latex]<\/p>\n

                      \n
                    1. Sketch the graph of [latex]f[\/latex].<\/li>\n
                    2. Is it possible to find a value [latex]k[\/latex] such that [latex]f(1)=k[\/latex], which makes [latex]f(x)[\/latex] continuous for all real numbers? Briefly explain.<\/li>\n<\/ol>\n
                      Show Solution<\/span><\/p>\n
                      \n

                      a.<\/p>\n

                      \"A<\/span>
                      \nb. It is not possible to redefine [latex]f(1)[\/latex] since the discontinuity is a jump discontinuity.<\/div>\n<\/div>\n<\/div>\n<\/div>\n

                      \n
                      \n

                      26.\u00a0<\/strong>Let [latex]f(x)=\\dfrac{x^4-1}{x^2-1}[\/latex] for [latex]x\\ne -1,1[\/latex].<\/p>\n

                        \n
                      1. Sketch the graph of [latex]f[\/latex].<\/li>\n
                      2. Is it possible to find values [latex]k_1[\/latex] and [latex]k_2[\/latex] such that [latex]f(-1)=k_1[\/latex] and [latex]f(1)=k_2[\/latex], and that makes [latex]f(x)[\/latex] continuous for all real numbers? Briefly explain.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n
                        \n
                        \n

                        27.\u00a0<\/strong>Sketch the graph of the function [latex]y=f(x)[\/latex] with properties 1 through 7.<\/p>\n

                          \n
                        1. The domain of [latex]f[\/latex] is [latex](\u2212\\infty,+\\infty)[\/latex].<\/li>\n
                        2. [latex]f[\/latex] has an infinite discontinuity at [latex]x=-6[\/latex].<\/li>\n
                        3. [latex]f(-6)=3[\/latex]<\/li>\n
                        4. [latex]\\underset{x\\to -3^-}{\\lim}f(x)=\\underset{x\\to -3^+}{\\lim}f(x)=2[\/latex]<\/li>\n
                        5. [latex]f(-3)=3[\/latex]<\/li>\n
                        6. [latex]f[\/latex] is left continuous but not right continuous at [latex]x=3[\/latex].<\/li>\n
                        7. [latex]\\underset{x\\to -\\infty}{\\lim}f(x)=\u2212\\infty[\/latex] and [latex]\\underset{x\\to +\\infty}{\\lim}f(x)=+\\infty[\/latex]<\/li>\n<\/ol>\n
                          Show Solution<\/span><\/p>\n
                          \n
                          \n

                          Answers may vary; see the following example:<\/p>\n

                          \"A<\/span><\/div>\n<\/div>\n<\/div>\n

                          <\/div>\n<\/div>\n<\/div>\n
                          \n
                          \n

                          28.\u00a0<\/strong>Sketch the graph of the function [latex]y=f(x)[\/latex] with properties 1 through 4.<\/p>\n

                            \n
                          1. The domain of [latex]f[\/latex] is [latex][0,5][\/latex].<\/li>\n
                          2. [latex]\\underset{x\\to 1^+}{\\lim}f(x)[\/latex] and [latex]\\underset{x\\to 1^-}{\\lim}f(x)[\/latex] exist and are equal.<\/li>\n
                          3. [latex]f(x)[\/latex] is left continuous but not continuous at [latex]x=2[\/latex], and right continuous but not continuous at [latex]x=3[\/latex].<\/li>\n
                          4. [latex]f(x)[\/latex] has a removable discontinuity at [latex]x=1[\/latex], a jump discontinuity at [latex]x=2[\/latex], and the following limits hold: [latex]\\underset{x\\to 3^-}{\\lim}f(x)=\u2212\\infty[\/latex] and [latex]\\underset{x\\to 3^+}{\\lim}f(x)=2[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n

                            In the following exercises (29-30), suppose [latex]y=f(x)[\/latex] is defined for all [latex]x[\/latex]. For each description, sketch a graph with the indicated property.<\/p>\n

                            \n
                            \n

                            29.\u00a0<\/strong>Discontinuous at [latex]x=1[\/latex] with [latex]\\underset{x\\to -1}{\\lim}f(x)=-1[\/latex] and [latex]\\underset{x\\to 2}{\\lim}f(x)=4[\/latex]<\/p>\n

                            Show Solution<\/span><\/p>\n
                            \n

                            Answers may vary; see the following example:<\/p>\n

                            \"The<\/span><\/div>\n<\/div>\n<\/div>\n<\/div>\n

                            \n
                            \n

                            30.\u00a0<\/strong>Discontinuous at [latex]x=2[\/latex] but continuous elsewhere with [latex]\\underset{x\\to 0}{\\lim}f(x)=\\frac{1}{2}[\/latex]<\/p>\n<\/div>\n<\/div>\n

                            Determine whether each of the given statements is true (31-37). Justify your responses with an explanation or counterexample.<\/p>\n

                            \n
                            \n

                            31.\u00a0<\/strong>[latex]f(t)=\\dfrac{2}{e^t-e^{-t}}[\/latex] is continuous everywhere.<\/p>\n

                            Show Solution<\/span><\/p>\n
                            \n

                            False. It is continuous over [latex](\u2212\\infty,0) \\cup (0,\\infty)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

                            \n
                            \n

                            32.\u00a0<\/strong>If the left- and right-hand limits of [latex]f(x)[\/latex] as [latex]x\\to a[\/latex] exist and are equal, then [latex]f[\/latex] cannot be discontinuous at [latex]x=a[\/latex].<\/p>\n<\/div>\n<\/div>\n

                            \n
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                            33.\u00a0<\/strong>If a function is not continuous at a point, then it is not defined at that point.<\/p>\n

                            Show Solution<\/span><\/p>\n
                            \n

                            False. Consider [latex]f(x)=\\begin{cases} x & \\text{ if } \\, x \\ne 0 \\\\ 4 & \\text{ if } \\, x = 0 \\end{cases}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

                            \n
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                            34.\u00a0<\/strong>According to the IVT, [latex]\\cos x - \\sin x - x = 2[\/latex] has a solution over the interval [latex][-1,1][\/latex].<\/p>\n<\/div>\n<\/div>\n

                            \n
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                            35.\u00a0<\/strong>If [latex]f(x)[\/latex] is continuous such that [latex]f(a)[\/latex] and [latex]f(b)[\/latex] have opposite signs, then [latex]f(x)=0[\/latex] has exactly one solution in [latex][a,b][\/latex].<\/p>\n

                            Show Solution<\/span><\/p>\n
                            \n

                            False. Consider [latex]f(x)= \\cos (x)[\/latex] on [latex][-\\pi, 2\\pi][\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

                            \n
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                            36.\u00a0<\/strong>The function [latex]f(x)=\\dfrac{x^2-4x+3}{x^2-1}[\/latex] is continuous over the interval [latex][0,3][\/latex].<\/p>\n<\/div>\n<\/div>\n

                            \n
                            \n

                            37.\u00a0<\/strong>If [latex]f(x)[\/latex] is continuous everywhere and [latex]f(a), f(b)>0[\/latex], then there is no root of [latex]f(x)[\/latex] in the interval [latex][a,b][\/latex].<\/p>\n

                            Show Solution<\/span><\/p>\n
                            \n

                            False. The IVT does not<\/em> work in reverse! Consider [latex](x-1)^2[\/latex] over the interval [latex][-2,2][\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

                            The following problems (38-39) consider the scalar form of Coulomb\u2019s law, which describes the electrostatic force between two point charges, such as electrons. It is given by the equation [latex]F(r)=k_e\\dfrac{|q_1q_2|}{r^2}[\/latex], where [latex]k_e[\/latex] is Coulomb\u2019s constant, [latex]q_i[\/latex] are the magnitudes of the charges of the two particles, and [latex]r[\/latex] is the distance between the two particles.<\/p>\n

                            \n
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                            38. [T]<\/strong> To simplify the calculation of a model with many interacting particles, after some threshold value [latex]r=R[\/latex], we approximate [latex]F[\/latex] as zero.<\/p>\n

                              \n
                            1. Explain the physical reasoning behind this assumption.<\/li>\n
                            2. What is the force equation?<\/li>\n
                            3. Evaluate the force [latex]F[\/latex] using both Coulomb\u2019s law and our approximation, assuming two protons with a charge magnitude of [latex]1.6022 \\times 10^{-19} \\, \\text{coulombs (C)}[\/latex], and the Coulomb constant [latex]k_e = 8.988 \\times 10^9 \\, \\text{Nm}^2\/\\text{C}^2[\/latex] are 1 m apart. Also, assume [latex]R<1\\text{m}[\/latex]. How much inaccuracy does our approximation generate? Is our approximation reasonable?<\/li>\n
                            4. Is there any finite value of [latex]R[\/latex] for which this system remains continuous at [latex]R[\/latex]?<\/li>\n<\/ol>\n<\/div>\n<\/div>\n
                              \n
                              \n

                              39. [T]\u00a0<\/strong>Instead of making the force 0 at [latex]R[\/latex], instead we let the force be [latex]10^{-20}[\/latex]\u00a0for [latex]r\\ge R[\/latex]. Assume two protons, which have a magnitude of charge [latex]1.6022 \\times 10^{-19} \\, \\text{C}[\/latex], and the Coulomb constant [latex]k_e=8.988 \\times 10^9 \\, \\text{Nm}^2\/\\text{C}^2[\/latex]. Is there a value [latex]R[\/latex] that can make this system continuous? If so, find it.<\/p>\n

                              Show Solution<\/span><\/p>\n
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                              [latex]R=0.0001519 \\, \\text{m}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

                              Recall the discussion on spacecraft from the Why It Matters. The following problems (40-42) consider a rocket launch from Earth\u2019s surface. The force of gravity on the rocket is given by [latex]F(d)=\\frac{-mk}{d^2}[\/latex], where [latex]m[\/latex] is the mass of the rocket, [latex]d[\/latex] is the distance of the rocket from the center of Earth, and [latex]k[\/latex] is a constant.<\/p>\n

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                              40. [T]<\/strong> Determine the value and units of [latex]k[\/latex] given that the mass of the rocket on Earth is 3 million kg. (Hint<\/em>: The distance from the center of Earth to its surface is 6378 km.)<\/p>\n<\/div>\n<\/div>\n

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                              41. [T]<\/strong> After a certain distance [latex]D[\/latex] has passed, the gravitational effect of Earth becomes quite negligible, so we can approximate the force function by [latex]F(d)=\\begin{cases} -\\dfrac{mk}{d^2} & \\text{ if } \\, d < D \\\\ 10,000 & \\text{ if } \\, d \\ge D \\end{cases}[\/latex] Find the necessary condition [latex]D[\/latex] such that the force function remains continuous.<\/p>\n

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                              [latex]D=63.78[\/latex] km<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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                              42.\u00a0<\/strong>As the rocket travels away from Earth\u2019s surface, there is a distance [latex]D[\/latex] where the rocket sheds some of its mass, since it no longer needs the excess fuel storage. We can write this function as [latex]F(d)=\\begin{cases} -\\dfrac{m_1 k}{d^2} & \\text{ if } \\, d < D \\\\ -\\dfrac{m_2 k}{d^2} & \\text{ if } \\, d \\ge D \\end{cases}[\/latex] Is there a [latex]D[\/latex]\u00a0value such that this function is continuous, assuming [latex]m_1 \\ne m_2[\/latex]?<\/p>\n<\/div>\n<\/div>\n

                              Prove the following functions are continuous everywhere (43-45).<\/p>\n

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                              43.\u00a0<\/strong>[latex]f(\\theta) = \\sin \\theta[\/latex]<\/p>\n

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                              For all values of [latex]a, \\, f(a)[\/latex] is defined, [latex]\\underset{\\theta \\to a}{\\lim}f(\\theta)[\/latex] exists, and [latex]\\underset{\\theta \\to a}{\\lim}f(\\theta)=f(a)[\/latex]. Therefore, [latex]f(\\theta)[\/latex] is continuous everywhere.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

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                              44.\u00a0<\/strong>[latex]g(x)=|x|[\/latex]<\/p>\n<\/div>\n<\/div>\n

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                              45.\u00a0<\/strong>Where is [latex]f(x)=\\begin{cases} 0 & \\text{ if } \\, x \\, \\text{is irrational} \\\\ 1 & \\text{ if } \\, x \\, \\text{is rational} \\end{cases}[\/latex] continuous?<\/p>\n

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                              Nowhere<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n\n\t\t\t

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