Final Practice Answers

These are the answers to the Final Practice.

P-Level

    1. \(\displaystyle\ln(10)10^x\)
    2. \(\displaystyle\frac{1}{x^2+1}\)
    3. \(\displaystyle\frac{3^x}{(x-1)^2+1}+\arctan(x-1)\ln(3)3^x+2x\mathrm{e}^{x^2}+\frac{\cos(x)+x\sin(x)+1}{(\cos(x)+1)^2}\)
    4. \(\displaystyle\frac{1}{x-2}+\sec^2(x-1)+\frac{10x\sin(x)-(5x^2-81)\cos(x)}{\sin^2(x)}\)
    1. \(2x^2+C\)
    2. \(\frac{1}{3}x^3+C\)
    3. \(\frac{1}{5}\tan(5x)+C\)
    4. \(\displaystyle\frac{3^{5x}}{5\ln(3)}+C\)
    5. \(4\arcsin(x)-5\sin(x)+\pi^2x+C\)
    6. \(\pi^2\arctan(x)-3\sec(x)+\frac{1}{4}\mathrm{e}^{4x}+C\)

  3. increasing on \((-\infty,3]\) and decreasing on \([3,\infty)\)

  4. convex on \((-\infty,0],[4,\infty)\) and concave on \([0,4]\)

  5. \(f^{\prime\prime}\) is positive on \((-\infty,2)\cup (4.5,8)\) and negative on \((2,4.5)\cup(8,\infty).\)

  6. \(\frac{\pi}{4}\)

  7. \(2x\sin(x^2)\cos(x^2)\)

  8. \(f'\) is positive on \((-3,-2)\) and \(f'\) is negative on \((-4,-3)\cup(-2,1)\cup(1,2)\)

  9. \(f\) is differentiable on \((-5,1)\cup(1,2)\cup(2,3)\cup(3,4)\cup(4,5)\)

  10. There many answers for \(I\) and \(J\). Here are two examples for each one: \(I=[-1,0]\), \(I=[-2,1]\), \(J=[2,4]\), \(J=[1,5]\).

  11. \(\omega(h)=2h\).

  12. By the intermediate value theorem, the following intervals will contain a zero of \(f\): \((-2,-1)\), \((-1,0)\) and \((3,4)\).

    1. \(\frac{19}{2}\)
    2. \(\frac{27}{2}\)
    3. \(-\frac{7}{6}\)
    4. \(6\)
    5. \(4+3\mathrm{e}^3\)

C-Level

  1. \(\displaystyle L_{-\frac{\pi}{2}}(x)=x+\frac{\pi}{2}\)

  2. \(\displaystyle L_{3}(x)=\frac{3}{8}x+\frac{23}{8}\)

  3. \(L_1(x)= (2+2\ln(2))(x-1)+2\)

    1. \(\displaystyle\frac{\ln(x)}{\ln(2)x}+\frac{\log_2(x)}{x}+\cos(x-1)+\frac{(10x-\ln(4)4^x)\sec(x)-(5x^2-4^x)\sec(x)\tan(x)}{(\sec(x))^2}+2xg'(x^2)\)
    2. \(\displaystyle-\frac{2\cos(x)}{\sqrt{1-(2x-4)^2}}-\arccos(2x-4)\sin(x)+4^{x^2+3x}\ln(4)\cdot(2x+3)+\frac{g'(x)}{2\sqrt{g(x)-1}}\)

  5. Take the derivative with respect to \(x\) of the equation and solve for \(\frac{\mathrm{d}y}{\mathrm{d}x}\) to obtain the following answers.

    1. \(\frac{\csc(x)\cot(x)-\sec^2(x)}{1+\sin(y-1)}\)
    2. \(\frac{\csc^2(x)-\tan(y)-1}{x\sec^2(y)+\sin(y-1)}\)

  6. Use linearity of antidifferentiation and reverse chain rule or u-substitution to get the following answers.

    1. \(\displaystyle2\sqrt{x^2+1}+\frac{3}{20}(5x-4)^4+C\)
    2. \(\displaystyle\ln|x^2+1|-\frac{1}{2}\mathrm{e}^{-x^2}+C\)
    3. \(\displaystyle\arctan(\mathrm{e}^x)+C\)
    4. \(\displaystyle\ln\left|10x^4+\frac{5}{2}x^2+x+20\right|+C\)
    5. \(\displaystyle -\frac{1}{\mathrm{e}^x+2^x}+C\)

  7. Use linearity of antidifferentiation and the fundamental theorem of calculus to get the following answers.

    1. \(\sin(8)-12\)
    2. \(14-4\ln(3)\)

  8. Sketch \(f^\prime\) and determine its sign to obtain that \(f\) is increasing on \([1,\infty)\) and decreasing on \((-\infty,1]\). As a result, it has a local min \(x=1\) and no local max.

  9. Sketch \(f^\prime\) and determine its sign to obtain that \(f\) is increasing on \([0,\infty)\) and decreasing on \((-\infty,0]\). As a result, it has a local min \(x=0\) and no local max.

  10. Sketch \(f^\prime\) and determine its sign to obtain that \(f\) is increasing on \((-\infty,0]\) and \([\ln(10),\infty)\), and decreasing on \([0,\ln(10)]\). As a result, it has a local max at \(0\) and a local min at \(\ln(10)\).

  11. Use \(V(t)=\tfrac{4}{3}\pi (r(t))^3\) and \(\frac{\mathrm{d}V}{\mathrm{d}t}=125\frac{\mathrm{cm}^3}{s}\) to obtain that \(\frac{\mathrm{d}r}{\mathrm{d}t}\Big|_{r(t)=15}=\frac{5}{36\pi}\frac{\mathrm{cm}}{s},\)

    1. \(\{-3,-2,1\}\)
    2. \([-3,-2]\)
    3. \([-4,-3]\cup[-2,1]\cup[1,2]\)

  13. local maxes at \(x=-2\) and \(x=4\) and local min \(x=0\)

    1. increasing on \([-0.5,0.5]\) and \([4,5]\); decreasing on \([-2,-0.5]\) and \([0.5,4]\).
    2. convex on \([-1.5,0]\) and \([2.5,5]\); concave on \([-2,-1.5]\) and \([0,2.5]\).
    3. local maximum at \(0.5\) and local minimum at \(-0.5\) and \(4\).
    4. inflection points at \(-1.5\), \(0\) and \(2.5\)

  15. Reconstruct the path by using \(c'(t)=v(t)\) and \(c(0)=(1,2)\) to obtain that \(c(t)=\Big(\tfrac{1}{3}t^3+t+1 2\sin(t)+\tfrac{4}{3}(t^2+1)^\frac{3}{2}+\tfrac{2}{3}\Big).\)

B-Level

  1. Use \(A(t)=4\pi (r(t))^2\) and \(\frac{\mathrm{d}V}{\mathrm{d}t}=4\frac{\mathrm{cm}^3}{s}\) to obtain that \(\frac{\mathrm{d}A}{\mathrm{d}t}\Big|_{r(t)=12\mathrm{in}}=\frac{2}{3}\frac{\mathrm{in}^2}{s}.\)

  2. Use Newton’s Method on \(f(x)=x^2-19\) with \(x_0\) to obtain the following approximations: \[x_0=4,\quad x_1=4.375,\quad x_2\approx 4.358928,\quad\text{and}\quad x_3\approx 4.358898.\]

  3. Take the derivative of \(f\) to identify the critical values of \(f\) and use that determine the local maximums and minimums of \(f\). Use the critical values and the endpoints of the interval to determine the global maximum and minimum.

    • local maximum \(x=-3\)
    • local minimum \(x=2\)
    • global maximum \(x=-3\)
    • global minimum \(x=2\)

  4. Take the derivative of \(f\) to identify the critical values of \(f\) and use that determine the local maximums and minimums of \(f\). Use the critical values and the endpoints of the interval to determine the global maximum and minimum.

    • no local max
    • no local min
    • global maximum \(x=-5\)
    • global minimum \(x=4\)

  5. Take the derivative of \(f\) to identify the critical values of \(f\) and use that determine the local maximums and minimums of \(f\). Use the critical values and the endpoints of the interval to determine the global maximum and minimum.

    • local max \(x=0\), \(x=\frac{4}{3}+\frac{\sqrt{10}}{3}\)
    • local min \(x=\frac{4}{3}-\frac{\sqrt{10}}{3}\)
    • global maximum \(x=\frac{4}{3}+\frac{\sqrt{10}}{3}\)
    • global minimum \(x=-1\)
    1. \(0\)
    2. \(1\)
    3. \(\frac{9}{4}\)
    4. \(\mathrm{e}^\frac{1}{15}\)

  7. Reconstruct the path by using \(c'(t)=v(t)\), \(v'(t)=a(t)\), \(c(0)=(1,2)\), and \(v(0)=\langle 2,4\rangle\) to obtain that \(c(t)=\left( \tfrac{1}{12}t^4+\tfrac{1}{6}t^3+\tfrac{1}{2}t^2+2t+1,\tfrac{1}{4}\mathrm{e}^{2t}-\sin(t)+\tfrac{9}{2}t+\tfrac{7}{4}\right).\)

A-Level

  1. The regions are shown below. The smallest intervals are given below
    1. \([-0.5,4.5]\)
    2. \([0.5,3.5]\)
    3. \([-0.5,4.5]\)
    4. \([1.25,2.75]\)

  1. Use the second order and first order information to identify where \(f\) is convex, concave, increasing and decreasing to produce a sketch like this.

    1. Use the graph and the formula for \(f\) to get that the limit is \(7\).
    2. Use the graph and the formula for \(f\) to get that the limit is \(\frac{5}{2}\).
    3. Use the sum rule for differentiation to get that the derivative evaluated at \(1\) is \(6\),
    4. Use the product rule for differentiation to get that the derivative evaluated at \(4\) is \(-\frac{21}{2}\).
    5. Use linearity of antidifferentiation to get that definite integral is \(71-\frac{1}{2}\pi\).
    6. Use linearity of antidifferentiation to get that definite integral is \(\frac{85}{4}\).
    7. Use L’Hopital’s rule to get that the limit is \(-4\).

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