Chapter 6.3 Practice
Questions
Determine all points at which each function \(f\) below can potentially attain a local maximum or a local minimum.
- \(f(x) = -2x^2+4\)
- \(f(x)=|5x-10|\)
- \(f(x)=\frac{x-4}{x^2-25}\)
- \(f(x)=\frac{x-4}{x^2+25}\)
- \(f(x)=\cos(x)+\frac{x}{2}\)
- \(f(x)=\sqrt[3]{x-2}+x\)
- \(f(x)=\frac{1}{3}x^3-\frac{3}{2}x^2+10x-5\)
- \(f(x)=(x+5)(x-3)(x-2)\)
- \(f(x)=|x-1|e^{-x^2}\)
For each function \(f\) and interval \(I\) given, determine the maximum and minimum values that \(f\) attains on \(I.\)
- \(f(x) = x^2+9\), \(I=[-1,3]\)
- \(f(x)=x^3+x^2-x\), \(I=\left[-1,\frac{1}{3}\right]\)
- \(f(x)=|5x-10|\), \(I=[-4,4]\)
- \(f(x)=\frac{1}{x^2+1}\), \(I=[-4,4]\)
- \(f(x)=\cos(x)+\frac{x}{2}\), \(I=\left[0,\frac{\pi}{2}\right]\)
- \(f(x)=\sqrt{x-4}\), \(I=\left[4,10\right]\)
- \(f(x)=x^2+\frac{16}{x}\), \(I=[1,5]\)
- \(f(x)=|x-4|e^{-x^2}\), \(I=[-2,6]\)
- \(f(x)=\frac{1}{3}x^3-\frac{1}{2}x^2-6x+7\), \(I=[-3,4]\)
- \(f(x)=(x-2)(x+5)(x-3)\), \(I=[-4,4]\)
For each function \(f\) and interval \(I\), determine whether \(f\) satisfies the conditions of Rolle’s Theorem on the given interval. If the function does satisfy Rolle’s theorem, find all numbers \(c\) that satisfy the conclusion of Rolle’s Theorem.
- \(f(x)=\cos(x)+\sin(x)\), \(I=\left[\frac{\pi}{4},\frac{3\pi}{4}\right]\)
- \(f(x)=\cos(x)+\sin(x)\), \(I=\left[-\frac{\pi}{4},\frac{3\pi}{4}\right]\)
- \(f(x)=-\sqrt{x}+\frac{1}{4}x\), \(I=\left[0,16\right]\)
- \(f(x)=x^3-x^2-2x+4\), \(I=\left[0,2\right]\)
- \(f(x)=|x-4|\), \(I=[0,8]\)
- \(f(x)=\frac{1}{(x-2)^2}\), \(I=[1,3]\)
For each function \(f\) and interval \(I\), determine whether \(f\) satisfies the conditions of the Mean Value Theorem on the given interval. If the function does satisfy the Mean Value Theorem, find all numbers \(c\) that satisfy the conclusion of the Mean Value Theorem.
- \(f(x)=3x^2+6x+2\), \(I=\left[-1,2\right]\)
- \(f(x)=\frac{x}{x-2}\), \(I=\left[1,4\right]\)
- \(f(x)=\frac{x}{x-2}\), \(I=\left[3,6\right]\) D. \(f(x)=|x-4|\), \(I=[0,8]\)
For each function \(f\) that is differentiable on \([1,4]\), \(x_0\) in \([1,4]\), and upper bound for \(|f'|\), if \(f(x_0)=-2\), then determine the smallest range of values that is guaranteed to contain \(f([1,4])\). Sketch the region as well.
- \(x_0=1\), \(|f'|\leq 3\)
- \(x_0=2\), \(|f'|\leq 3\)
- \(x_0=3\), \(|f'|\leq 3\)
- \(x_0=1\), \(|f'|\leq \frac{1}{2}\)
- \(x_0=2\), \(|f'|\leq \frac{1}{2}\)
- \(x_0=3\), \(|f'|\leq \frac{1}{2}\)
For each function \(f\) that is differentiable on the interval \(I\), and for \(f\) with the given properties, determine the smallest range of values that is guaranteed to contain \(f(I).\) Sketch the region as well.
- \(I=[1,4]\), \(f(1)=3\), \(f(4)=7\), \(|f'|\leq 2\)
- \(I=[-3,5]\), \(f(-3)=2\), \(f(5)=4\), \(|f'|\leq \frac{1}{4}\)
Prove the following statements using the Mean Value Theorem or Rolle’s Theorem.
- The equation \(\sin(x)-x+1=0\) has exactly one real root.
- If \(f\) is a differentiable function so that \(2\leq f'(x)\leq 4\), then \(10\leq f(6)-f(1)\leq 20.\)
Follow the subsequent parts to show that \(e^{x}>1+x\) for all \(x>0\). Take \(g(x)=e^x-(1+x).\)
- Verify that \(g'\) is positive for all \(x>0\).
- Verify that \(g\) satisfies all the conditions of the Mean Value Theorem on \([0,b]\), where \(b\) is greater than 0.
- Apply the mean value theorem on \([0,b]\) to conclude that \(e^b>1+b.\)
Take \(f\) and \(g\) to be differentiable functions on the interval \([10,18]\) so that \(f(10)=10\), \(g(10)=-8\), and that for any \(x\) in \((10,18),\) \[f'(x)-g'(x)=0.\] Determine \(f(18)-f(18).\)
For each function \(f\) that is given below, determine the antiderivative of \(f:\)
- \(f(x) = \mathrm{e}+\cos(x)\)
- \(f(x) = \frac{50}{1+x^2}\)
- \(f(x) = \frac{1}{x}-\frac{1}{\sqrt{x}}\)
- \(f(x)= \frac{x^{4/3}-x+\ln(2)}{x^2}\)
- \(f(x)= \frac{1}{\sin^2(x)}\)
- \(f(x)=2^x+3x\)
- \(f(x)=3g(x)-\sec(x)\tan(x)\) where \(g\) is function such that an antiderivative of \(g\) \(G(x)=\mathrm{e}^x\sin(x)+1\)
- \(f(x)=4e^x-x^4\)
- \(f(x)=\sin(x)+\mathrm{e}^2-\cos(x)\)
- \(f(x)=\frac{\mathrm{e}^{2x}-5\mathrm{e}^x}{\mathrm{e}^x}\)
- \(f(x)=\sec^2(x)+\ln(2)\)
The function \(F\) is an antiderivative of \(f\), where \[F(x)=\sec(x)8^x.\] Determine \[\int f(x)\,\mathrm{d}x.\]
For each function \(f\) with the given property, identify the function \(f\).
- The function \(f\) has the property that \(\displaystyle \int f(x)\,\mathrm{d}x=\frac{5^x}{\log_5(x)}+C\)
- The function \(f\) has the property that \(\displaystyle \int f(x)\,\mathrm{d}x=\log_8(x)+\sin(x)+C.\)
- The function \(f\) has the property that \(\displaystyle \int f(x)\,\mathrm{d}x=\cos(\sin(x))+C.\)
- The function \(f\) has the property that \(\displaystyle \int f(x)\,\mathrm{d}x=\log_7(x)\arctan(x)+C.\)
Calculate the following:
- \(\displaystyle\int_{2}^4x^2\,\mathrm{d}x\)
- \(\displaystyle\int_{0}^1 \frac{50}{1+x^2}\,\mathrm{d}x\)
Take \(g\) to be an integrable function whose graph is given below. Calculate the following indefinite integrals.
- \(\displaystyle \int_{0}^{4} g(x)\,\mathrm{d}x\)
- \(\displaystyle \int_{0}^{2} g(x)\,\mathrm{d}x\)
- \(\displaystyle \int_{0}^{2} g(2x)\,\mathrm{d}x\)
- \(\displaystyle \int_{0}^{8} g(x)\,\mathrm{d}x\)
Determine the following indefinite integrals:
- \(\displaystyle\int \sin(x-4)\,\mathrm{d}x\)
- \(\displaystyle\int \frac{\sqrt{\arctan(x)}}{4+4x^2}\,\mathrm{d}x\)
- \(\displaystyle\int (2x-4)\csc^2(x^2-4x)\,\mathrm{d}x\)
- \(\displaystyle \int \mathrm{e}^{\sin(x)}\cos(x)\,\mathrm{d}x\)
- \(\displaystyle \int \cos\left(\frac{2^x}{\ln(2)}\right)2^x\,\mathrm{d}x\)
- \(\displaystyle \int -\frac{\cos(x)}{1+\sin^2(x)}\,\mathrm{d}x\)
- \(\displaystyle \int \mathrm{e}^{x-2}\,dx\)
- \(\displaystyle \int 2^x\left(2^x+\cos(1)\right)^{4/3}\,dx\)
- \(\displaystyle \int \cos(x)\sec^2(2\sin(x))\,dx\)
- \(\displaystyle \int \frac{9x-9}{x^2-2x}\,dx\)
- \(\displaystyle \int \frac{\arctan(x)}{x^2+1}\,dx\)
- \(\displaystyle \int \frac{3^{1/x}}{x^2}\,dx\)
- \(\displaystyle \int \frac{x^3}{\sqrt{x^2+3}}\,dx\)
- \(\displaystyle \int \sqrt{f(x)}\,f'(x)\,dx\) where \(f\) is a differentiable function
- \(\displaystyle \int \frac{g\left(\sqrt{x+1}\right)}{\sqrt{x+1}}\,dx\) where \(g\) is a function so that an antiderivative of \(g\) is \(\sin(x^2+x)\)
- \(\displaystyle \int \frac{24x^2+12x+4}{1+(8x^3+6x^2+4x+6)^2}\,\mathrm{d}x\)
- \(\displaystyle \int -\frac{csc^2(x)}{\cot(x)}\,\mathrm{d}x\)
Take \(f\) and \(g\) to be functions with the property that \(F\) is an antiderivative of \(f\), and \(G\) is an antiderivative of \(g\), where \[F(x)=e^{\cot(x)}\quad\text{and}\quad G(x)=3^x+\log_3(x).\] Determine the indefinite integral \[\int (-9f(x)-2g(x))\,\mathrm{d}x.\]
Take \(f\) to be a function with the property that \(F\) is an antiderivative of \(f\), where \[F(x)=\ln(\sec(x)).\] Determine the indefinite integral \[\int -7f(-9x)\,\mathrm{d}x.\]
Take \(f\) and \(g\) to be functions with the property that \(F\) is an antiderivative of \(f\), and \(G\) is an antiderivative of \(g\), where \[F(x)=\frac{\log_6(x)}{\tan(x)}\quad\text{and}\quad G(x)=\frac{\arctan(x)}{\cot(x)}.\] Determine the indefinite integral \[\int (g(x)f(G(x)))\,\mathrm{d}x.\]
Take \(f\) and \(g\) to be functions with the property that \(F\) is an antiderivative of \(f\), and \(G\) is an antiderivative of \(g\), where \[F(x)=\mathrm{e}^x\cos(x)\quad\text{and}\quad G(x)=5^x+\tan(x).\] Determine the indefinite integral \[\int (3g(5x)f(G(5x)))\,\mathrm{d}x.\]
Use the fact that \(\left(x\frac{2^x}{\ln(2)}\right)'=\frac{2^x}{\ln(2)}+x2^x\) to determine \[\int x2^x\,\mathrm{d}x.\]
Verify that the following are of indeterminate forms and use L’Hospital’s rule to determine the given limits:
- \(\displaystyle\lim_{x\to 0}\frac{x^3+x}{x^2+x}\)
- \(\displaystyle\lim_{x\to 0^+}\frac{x^2+x}{x^3+x}\)
- \(\displaystyle\lim_{x\to 2^+} \frac{\ln(x-1)}{x-2}\)
- \(\displaystyle\lim_{x\to \infty} \frac{\mathrm{e}^{x}}{x}\)
- \(\displaystyle\lim_{x\to \infty}\frac{x^5+4x^2-3x}{x^6+x^3+x^2+1}\)
- \(\displaystyle\lim_{x\to \infty}\frac{3x^5+3}{4x^5-3x^2}\)
- \(\displaystyle\lim_{x\to 3^+}\left(x-3\right)^{x-3}\)
- \(\displaystyle\lim_{x\to 3^+}(x-2)^{\frac{1}{x-3}}\)
- \(\displaystyle\lim_{x\to 0^+}\sqrt{x}\cdot \mathrm{e}^{\frac{1}{x}}\)
- \(\displaystyle\lim_{x\to -\infty}x^3e^{x}\)
- \(\displaystyle\lim_{x\to 0}\left(\frac{1}{x^2}\right)^{x}\)
- \(\displaystyle\lim_{x\to \infty}\sqrt{x^2+2x}-x\)
- \(\displaystyle\lim_{x\to \infty}\sqrt{x^2+1}-x^2\)
- \(\displaystyle\lim_{x\to 0^+}\left(\frac{x+1}{x}-\frac{1}{\ln(x+1)}\right)\)
- \(\displaystyle\lim_{x\to 0}\left(1+3x\right)^{1/x}\)
- \(\displaystyle\lim_{x\to 0^+}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x}}\)
- \(\displaystyle \lim_{x\to\infty}\left((x^5+8x^4+50)^\frac{1}{5}-x\right)\)
- \(\displaystyle\lim_{x\to\infty} (\ln(x+3)-\ln(x+1))\)
- \(\displaystyle\lim_{x\to 0^+} x^{x\sin(x)}\)
Answers
- \(1\)
- \(2\)
- None
- \(4-\sqrt{41},4+\sqrt{41}\)
- \(\frac{\pi}{6}+2\pi k, \frac{5\pi}{6}+2\pi k\) where \(k\) is an integer
- None
- \(\frac{\sqrt{288}}{6},-\frac{\sqrt{288}}{6}\)
- \(\frac{1-\sqrt{3}}{2},1,\frac{1+\sqrt{3}}{2}\)
- maximum value of \(18\), minimum value of \(9\)
- maximum value of \(1\), minimum value of \(-\frac{5}{27}\)
- maximum value of \(30\), minimum value of \(0\)
- maximum value of \(1\), minimum value of \(\frac{1}{17}\)
- maximum value of \(\frac{\pi}{2}\), minimum value of \(1\)
- maximum value of \(\sqrt{6}\), minimum value of \(0\)
- maximum value of \(\frac{141}{5}\), minimum value of \(12\)
- maximum value of \(|-2-\frac{\sqrt{18}}{2}|\mathrm{e}^{-\left(\frac{4-\sqrt{18}}{2}\right)^2}\), minimum value of \(0\)
- maximum value of \(\frac{43}{3}\), minimum value of \(-\frac{13}{2}\)
- maximum value of \(\left(-\sqrt{19}{3}-2\right)\left(-\sqrt{19}{3}+5\right)\left(-\sqrt{19}{3}-3\right)\), minimum value of \(\left(\sqrt{19}{3}-2\right)\left(\sqrt{19}{3}+5\right)\left(\sqrt{19}{3}-3\right)\)
- No, because \(f\left(\frac{\pi}{4}\right)\) does not equal \(f\left(\frac{3\pi}{4}\right)\)
- Yes, \(c=\frac{\pi}{4}\)
- Yes, \(c=4\)
- Yes, \(c=\frac{1+\sqrt{7}}{3}\)
- No, because \(f\) is not differentiable at \(x=4\)
- No, because \(f\) is not differentiable at \(2\).
- Yes, \(c=\frac{1}{2}\)
- No, not continuous at \(x=2\)
- Yes, \(c=4\)
- No, not differentiable at \(x=4\)
- \([-11,7]\)
- \([-8,4]\)
- \([-8,4]\)
- \([-3.5,-0.5]\)
- \([-3,-1]\)
- \([-3,-1]\)
- \([2,8]\)
- \([0,4]\)
- Define \(f(x)=\sin(x)-x+1\). First use the Intermediate Value Theorem to conclude \(f\) has a root in the interval \([1,3].\) Use the Rolle’s theorem to conclude there is only one root.
- Because \(f\) is a differentiable function on \(\mathbb{R}\), it is also continuous on \(\mathbb{R}\). Apply the Mean Value Theorem on \(f\) on the interval \([1,6]\) to get \(f'(c)=\frac{f(6)-f(1)}{5}\) for some real number \(c\) in \((1,6)\). Use the bound on the derivative to get the desired result.
- The derivative is \(g'(x)=\mathrm{e}^x-1\). When \(x>0\), \(\mathrm{e}^x>1\) so \(g'\) is positive.
- Verify \(g\) is continuous on \([0,b]\) and differentiable on \((0,b)\)
- Apply the Mean Value Theorem to \(g\) on \([0,b]\) to get \(g'(c)=\frac{g(b)-g(0)}{b-0}\) for some \(c\) in \((0,c)\). Since \(g'\) is positive, \(\frac{g(b)-g(0)}{b-0}>0\). Simplify and rewrite the expression to get the desired result.
\(18\)
- \(F(x)=\mathrm{e}x+\sin(x)+C\)
- \(F(x)=50\arctan(x)+C\)
- \(F(x)=\ln|x|-2\sqrt{x}+C\)
- \(F(x)= 3\sqrt[3]{x}-\ln|x|-\frac{\ln(2)}{x}+C\)
- \(F(x)= -\cot(x)+C\)
- \(F(x)=\frac{2^x}{\ln(2)}+\frac{3}{2}x^2+C\)
- \(F(x)=3\mathrm{e}^{x}\sin(x)+3-\sec(x)+C\)
- \(F(x)=4\mathrm{e}^x-\frac{1}{5}x^5+C\)
- \(F(x)=-\cos(x)+\mathrm{e}^2x-\sin(x)+C\)
- \(F(x)=\mathrm{e}^x-5x+C\)
- \(F(x)=\tan(x)+\ln(2)x+C\)
\(\int f(x)\,\mathrm{d}x=\sec(x)8^x+C\)
- \(f(x)=\frac{\ln(5)5^x\log_5(x)-\frac{5^x}{x\ln(5)}}{(\log_5(x))^2}\)
- \(f(x)=\frac{1}{x\ln(8)}+\cos(x)+C.\)
- \(f(x)=-\sin(\sin(x))\cos(x)+C\)
- \(f(x)=\frac{\arctan(x)}{x\ln(7)}+\frac{\log_7(x)}{x^2+1}\)
- \(\frac{56}{3}\)
- \(\frac{25}{2}\pi\)
- \(\frac{1}{2}\pi+2\)
- \(\frac{1}{2}\pi\)
- \(\frac{1}{4}\pi\) d.\(\frac{1}{2}\pi-5\)
- \(-\cos(x-4)+C\)
- \(\frac{1}{6}\left(\arctan(x)\right)^{\frac{3}{2}}+C\)
- \(-\cot(x^2-4x)+C\)
- \(\mathrm{e}^{\sin(x)}+C\)
- \(\sin\left(\frac{2^x}{\ln(2)}\right)+C\)
- \(-\arctan(\sin(x))+C\)
- \(e^{x-2}+C\)
- \(\frac{3}{7\ln(2)}(2^x+\cos(1))^{7/3}+C\)
- \(\frac{1}{2}\tan(2\sin(x))+C\)
- \(\frac{9}{2}\ln|x^2-2x|+C\)
- \(\frac{\arctan^2(x)}{2}+C\)
- \(-\frac{1}{\ln 3} 3^{1/x}+C\)
- \(\frac{1}{3}(x^2+3)^{3/2}-3\sqrt{x^2+3}+C\)
- \(\frac{1}{2}(f(x))^{3/2}+C\)
- \(2\sin\left(\left(\sqrt{x+1}\right)^2+\sqrt{x+1}\right)+C\)
- \(\arctan(8x^3+6x^2+4x+6)+C\)
- \(\ln|\cot(x)|+C\)
\(-9e^{\cot(x)}-2\cdot 3^x-2\log_3(x)+C\)
\(\frac{7}{9}\ln(\sec(-9x))+C\)
\(\frac{\log_6\left(\frac{\arctan(x)}{\cot(x)}\right)}{\tan\left(\frac{\arctan(x)}{\cot(x)}\right)}+C\)
\(\frac{3}{5}\exp(5^{5x}+\tan(5x)))\cos(5^{5x}+\tan(5x))\)
\(\frac{x2^x}{\ln(2)}-\frac{2^x}{(\ln(2))^2}+C\)
- \(0\)
- \(1\)
- \(1\)
- \(\infty\)
- \(0\)
- \(\frac{3}{4}\)
- \(1\)
- \(e\)
- \(\infty\)
- \(0\)
- \(1\)
- \(1\)
- \(-\infty\)
- \(\frac{1}{2}\)
- \(e^3\)
- \(1\)
- \(\frac{8}{5}\)
- \(0\)
- \(1\)