Practice Problems

Section 3.3

1. (a) Cont. at 0, 1, 2, 3, 5 (b) Diff. at 0, 3, 5

3.

$\begin{array}{c|c|c|c|c|l|l|l|l}\mathrm{x} & \mathrm{f}(\mathrm{x}) & \mathrm{f}^{\prime}(\mathrm{x}) & \mathrm{g}(\mathrm{x}) & \mathrm{g} '(\mathrm{x}) & \mathrm{f}(\mathrm{x}) \cdot \mathrm{g}(\mathrm{x}) & \mathrm{D}(\mathrm{f}(\mathrm{x}) \cdot \mathrm{g}(\mathrm{x})) & \mathrm{f}(\mathrm{x}) / \mathrm{g}(\mathrm{x}) & \mathrm{D}(\mathrm{f}(\mathrm{x}) / \mathrm{g}(\mathrm{x})) \\\hline 0 & 2 & 3 & 1 & 5 & \mathbf{2} & \mathbf{1 3} & \mathbf{2} & -7 \\1 & -3 & 2 & 5 & -2 & \mathbf{- 1 5} & \mathbf{1 6} & \mathbf{- 3} / \mathbf{5} & \mathbf{4} / \mathbf{2 5} \\2 & 0 & -3 & 2 & 4 & \mathbf{0} & -\mathbf{6} & \mathbf{0} & -\mathbf{3} / \mathbf{2} \\3 & 1 & -1 & 0 & 3 & \mathbf{0} & \mathbf{3} & \text { undef } & \text { undef }\end{array}$

5. 

$\begin{array}{l|c|l|c|c|c|l|l|l|l|l}\mathrm{x} & \mathrm{f}(\mathrm{x}) & \mathrm{f}^{\prime}(\mathrm{x}) & \mathrm{g}(\mathrm{x}) & \mathrm{g}^{\prime}(\mathrm{x}) & \mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x}) & \mathrm{f}(\mathrm{x}) \cdot \mathrm{g}(\mathrm{x}) & \mathrm{f}(\mathrm{x}) / \mathrm{g}(\mathrm{x}) & \mathrm{D}(\mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x})) & \mathbf{D}(\mathrm{f}(\mathrm{x}) \cdot \mathrm{g}(\mathrm{x})) & \mathrm{D}(\mathrm{f}(\mathrm{x}) / \mathrm{g}(\mathrm{x})) \\\hline 1 & 3 & -2 & 2 & 2 & 5 & 6 & 3 / 2 & 0 & 2 & -10 / 4 \\2 & 1 & 0 & 3 & 1 / 2 & 4 & 3 & 1 / 3 & 1 / 2 & 1 / 2 & -1 / 18 \\3 & 2 & 1 & 2 & -1 & 4 & 4 & 1 & 0 & 0 & 1\end{array}$

7. (a) $\quad D((x-5)(3 x+7))=(x-5) 3+(3 x+7) 1=6 x-8$

(b) $\quad$  $D\left(3 x^{2}-8 x-35\right)=6 x-8$, the same result as in (a)

9. $\frac{\mathrm{d}}{\mathrm{dx}} \frac{\cos (\mathrm{x})}{\mathrm{x}^{2}}=\frac{\mathrm{x}^{2}(-\sin (\mathrm{x}))-(\cos (\mathrm{x}))(2 \mathrm{x})}{\left(\mathrm{x}^{2}\right)^{2}}=-\frac{\mathrm{x} \sin (\mathrm{x})+2 \cos (\mathrm{x})}{\mathrm{x}^{3}}$

11. $D\left(\sin ^{2}(x)\right)=\sin (x) \cos (x)+\sin (x) \cos (x)=2 \sin (x) \cos (x)$, $D\left(\cos ^{2}(x)\right)=\cos (x)(-\sin (x))+\cos (x)(-\sin (x))=-2 \sin (x) \cos (x)$

13. $\mathrm{f}(\mathrm{x})=\mathrm{ax}^{2}+\mathrm{b} \mathrm{x}+\mathrm{c}$ so $\mathrm{f}(0)=\mathrm{c} .$ Then $\mathrm{f}(0)=0$ implies that $\mathrm{c}=0$ $f^{\prime}(x)=2 a x+b$ so $f^{\prime}(0)=b$ and $f^{\prime}(0)=0$ implies that $b=0$

Finally, $f^{\prime}(10)=20 a+b=20 a$ so $f^{\prime}(10)=30$ implies that $20 a=30$ and $a=3 / 2$. $f(x)=\frac{3}{2} x^{2}+0 x+0$ has $f(0)=0, f^{\prime}(0)=0$, and $f^{\prime}(10)=30$

15. Their graphs are vertical shifts of each other, and their derivatives are equal.

17. $\mathrm{f}(\mathrm{x}) \mathrm{g}(\mathrm{x})=\mathrm{k}$ so $\mathrm{D}(\mathrm{f}(\mathrm{x}) \mathrm{g}(\mathrm{x}))=\mathrm{D}(\mathrm{k})=0$ and $\mathrm{f}(\mathrm{x}) \mathrm{g}^{\prime}(\mathrm{x})+\mathrm{g}(\mathrm{x}) \mathrm{f}^{\prime}(\mathrm{x})=0$.

If $f(x) \neq 0$ and $g(x) \neq 0$, then $\quad \frac{f^{\prime}(x)}{f(x)}=-\frac{g^{\prime}(x)}{g(x)}$

19. $\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}-5$ so $\mathrm{f}^{\prime}(1)=-3 . \mathrm{f}^{\prime}(\mathrm{x})=0$ if $\mathrm{x}=5 / 2$.

21. $\mathrm{f}^{\prime}(\mathrm{x})=3+2 \sin (\mathrm{x})$ so $\mathrm{f}^{\prime}(1)=3+2 \sin (1) \approx 4.68 . \mathrm{f}^{\prime}(\mathrm{x})$ never equals 0 since $\sin (\mathrm{x})$ never equals $-3 / 2$.

23. $\mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}+18 \mathrm{x}=3 \mathrm{x}(\mathrm{x}+6)$ so $\mathrm{f}^{\prime}(1)=21 . \mathrm{f}^{\prime}(\mathrm{x})=0$ if $\mathrm{x}=0$ or $-6$.

25. $\mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}+4 \mathrm{x}+2$ so $\mathrm{f}^{\prime}(1)=9 . \mathrm{f}^{\prime}(\mathrm{x})=0$ for no values of $\mathrm{x}$ (the discriminant $\left.4^{2}-4(3)(2) < 0\right)$.

27. $f^{\prime}(x)=x^{\cdot} \cos (x)+\sin (x)$ so $f^{\prime}(1)=1 \cdot \cos (1)+\sin (1) \approx 1.38 .$ The graph of $f^{\prime}(x)$ crosses the $x-a x$ is infinitely often. The root of $f^{\prime}$ at $x=0$ is easy to see (and verify). Other roots of $f$ ', such as near $\mathrm{x}=2.03$ and $4.91$ and $-2.03$, can be found numerically using the Bisection algorithm or graphically using the "zoom" or "trace" features on some calculators.

29. $\mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}+2 \mathrm{Ax}+\mathrm{B} .$ The graph of $\mathrm{y}=\mathrm{f}(\mathrm{x})$ has two distinct "vertices" if $\mathrm{f}^{\prime}(\mathrm{x})=0$ for two distinct values of $x .$ This occurs if the discriminant of $3 x^{2}+2 A x+B$ is greater than $0:(2 A)^{2}-4(3)(B)>0$

31. Everywhere except at $\mathrm{x}=-3$.

33. Everywhere except at $\mathrm{x}=0$ and 3.

35. Everywhere except at $\mathrm{x}=1$

37. Everywhere. The only possible difficulty is at $\mathrm{x}=0$, and the definition of the derivative gives $\mathrm{f}^{\prime}(0)=1$. The derivatives of the "two pieces" of $\mathrm{f}$ match at $\mathrm{x}=0$ to give a differentiable function there.

39. Continuity of $\mathrm{f}$ at $\mathrm{x}=1$ requires $\mathrm{A}+\mathrm{B}=2 .$ The "left derivative" $\mathrm{f}$ at $\mathrm{x}=1$ is $\mathrm{D}(\mathrm{Ax}+\mathrm{B})=\mathrm{A}$ and the "right derivative" of $\mathrm{f}$ at $\mathrm{x}=1$ is $3\left(\right.$ if $\mathrm{x}>1$ then $\left.\mathrm{D}\left(\mathrm{x}^{2}+\mathrm{x}\right)=2 \mathrm{x}+1\right)$ so to achieve differentiability $A=3$ and $B=2-A=-1$

41. $\mathrm{h}(\mathrm{x})=128 \mathrm{x}-2.65 \mathrm{x}^{2} \mathrm{ft}$.

(a) $\mathrm{h}^{\prime}(\mathrm{x})=128-5.3 \mathrm{x}$ so $\mathrm{h}^{\prime}(0)=128 \mathrm{ft} / \mathrm{sec}, \mathrm{h}^{\prime}(1)=122.7 \mathrm{ft} / \mathrm{sec}$, and $\mathrm{h}^{\prime}(2)=117.4 \mathrm{ft} / \mathrm{sec}$.

(b) $v(x)=h^{\prime}(x)=128-5.3 \times \mathrm{ft} / \mathrm{sec}$.

(c) $\mathrm{v}(\mathrm{x})=0$ when $\mathrm{x}=128 / 5.3 \approx 24.15 \mathrm{sec}$.

(d) $\mathrm{h}(24.15) \approx 1,545.66 \mathrm{ft}$

(e) about $48.3$ seconds: $24.15$ up and $24.15$ down

43. $h(x)=v_{0} x-16 x^{2}$ ft.

(a) $\mathrm{h}^{\prime}(\mathrm{x})=\mathrm{v}_{\mathrm{o}}-32 \mathrm{x} \mathrm{ft} / \mathrm{sec}$

(b) Max height when $\mathrm{x}=\mathrm{v}_{\mathrm{o}} / 32:$ max height $=\mathrm{h}\left(\mathrm{v}_{\mathrm{o}} / 32\right)=\mathrm{v}_{\mathrm{o}}\left(\mathrm{v}_{\mathrm{o}} / 32\right)-16\left(\mathrm{v}_{\mathrm{o}} / 32\right)^{2}=\left(\mathrm{v}_{\mathrm{o}}\right)^{2} / 64$ feet.

(c) Time aloft $=2\left(\mathrm{v}_{\mathrm{o}} / 32\right)=\mathrm{v}_{\mathrm{o}} / 16$ seconds.

45. (a) $\left(\mathrm{v}_{\mathrm{o}}\right)^{2} / 64=6.5$, so $\mathrm{v}_{\mathrm{o}}=8 \sqrt{6.5} \approx 20.396 \mathrm{ft} / \mathrm{sec}$.

(b) $2\left(\mathrm{v}_{\mathrm{o}} / 32\right)=\frac{8 \sqrt{6.5}}{16} \approx 1.27$ seconds.

(c) Max height $=\frac{\left(\mathrm{v}_{\mathrm{o}}\right)^{2}}{2 \mathrm{~g}}=\frac{(8 \sqrt{6.5})^{2}}{2(5.3)}=\frac{416}{10.6} \approx 39.25$ feet.

47. (a) $y^{\prime}=-\frac{1}{x^{2}} ; y^{\prime}(2)=-1 / 4$ so $y-1 / 2=(-1 / 4)(x-2)$ or $y=(-1 / 4) x+1$

(b) $x$-intercept at $x=4, y$-intercept at $y=1$

(c) $\mathrm{A}=\frac{1}{2}($ base $)($ height $)=\frac{1}{2}(4)(1)=2$.

49. Since $(1,4)$ and $(3,14)$ are on the parabola, we need $a+b+c=4$ and $9 a+3 b+c=14$. Then, subtracting the first equation from the second, $8 a+2 b=10$.

$f^{\prime}(x) \equiv 2 a x+b$ so $f^{\prime}(3)=6 a+b=9$, the slope of $y=9 x-13 .$ Now solve the system $8 a+2 b=10$ and $6 a+b=9$ to get $a=2$ and $b=-3 .$ Then use $a+b+c=4$ to get $c=5 . a=2, b=-3, c=5$

51. (a) $f(x)=x^3$

(b) $g(x)=x^{3}+1$

(c) If $\mathrm{h}(\mathrm{x})=\mathrm{x}^{3}+\mathrm{C}$ for any constant $\mathrm{C}$, then $\mathrm{D}(\mathrm{h}(\mathrm{x}))=3 \mathrm{x}^{2}$.

53. (a) For $0 \leq x \leq 1, f^{\prime}(x)=1$ so $f(x)=x+C$. Since $f(0)=0$, we know $C=0$ and $f(x)=x$.

For $1 \leq x \leq 3, f^{\prime}(x)=2-x$ so $f(x)=2 x-\frac{1}{2} x^{2}+K .$ Since $f(1)=1$, we know $\mathrm{K}=-1 / 2$ and $\mathrm{f}(\mathrm{x})=2 \mathrm{x}-\frac{1}{2} \mathrm{x}^{2}-\frac{1}{2}$

For $3 \leq x \leq 4, f^{\prime}(x)=x-4$ so $f(x)=\frac{1}{2} x^{2}-4 x+L$. Since $f(3)=1$, we know $L=17 / 2$ and $f(x)=\frac{1}{2} x^{2}-4 x+\frac{17}{2}$

(b) This graph is a vertical shift, up 1 unit, of the graph in part (a).