Practice Problems

1. Local maximums at $\mathrm{x}=3$, $\mathrm{x}=5$, $\mathrm{x}=9$, and $\mathrm{x}=13$. Global maximums at $\mathrm{x}=3$ and $\mathrm{x}=13$.
Local minimums at $\mathrm{x}=1$, $\mathrm{x}=4.5$, $\mathrm{x}=7$, and $\mathrm{x}=10.5$. Global minimum at $\mathrm{x}=7$.

3. $f(x)=x^{2}+8 x+7$ so $f^{\prime}(x)=2 x+8$ which is defined for all values of $x$. $f^{\prime}(x)=0$ when $\mathrm{x}=-4$ so $\mathrm{x}=-4$ is a critical number. There are no endpoints.
The only critical number is $x=-4$, and the only critical point is $(-4, f(-4))=(-4,-9)$ which is the global (and local) minimum.

5. $f(x)=\sin (x)$ so $f^{\prime}(x)=\cos (x)$ which is defined for all values of $x$. $f^{\prime}(x)=0$ when $x=\frac{\pi}{2}+n \pi$ so the values $\mathrm{x}=\frac{\pi}{2}+\mathrm{n} \pi$ are critical numbers. There are no endpoints.
$\mathrm{f}(\mathrm{x})=\sin (\mathrm{x})$ has local and global maximums at $\mathrm{x}=\frac{\pi}{2}+2 \mathrm{n} \pi$, and global and local minimums at $\mathrm{x}=\frac{3 \pi}{2}+2 \mathrm{n} \pi$.

7. $\mathrm{f}(\mathrm{x})=(\mathrm{x}-1)^{2}(\mathrm{x}-3)$ so $\mathrm{f}^{\prime}(\mathrm{x})=(\mathrm{x}-1)^{2}+2(\mathrm{x}-1)(\mathrm{x}-3)=(\mathrm{x}-1)(3 \mathrm{x}-7)$ which is defined for all values of $x$. $f^{\prime}(x)=0$ when $x=1$ and $x=7 / 3$ so $x=1$ and $x=7 / 3$ are critical numbers. There are no endpoints. The only critical points are $(1,0)$ which is a local maximum and $(7 / 3,-32 / 27)$ which is a local minimum. When the interval is the entire real number line, this function does not have a global maximum or global minimum.

9. $\mathrm{f}(\mathrm{x})=2 \mathrm{x}^{3}-96 \mathrm{x}+42$ so $\mathrm{f}^{\prime}(\mathrm{x})=6 \mathrm{x}^{2}-96$ which is defined for all values of $\mathrm{x}$. $\mathrm{f}^{\prime}(\mathrm{x})=6(\mathrm{x}+4)(\mathrm{x}-4)=0$ when $\mathrm{x}=-4$ and $\mathrm{x}=4$ so $\mathrm{x}=-4$ and $\mathrm{x}=4$ are critical numbers. There are no endpoints. The only critical points are $(-4,298)$ which is a local maximum and $(4,-214)$ which is a local minimum. When the interval is the entire real number line, this function does not have a global maximum or global minimum.

11. $\mathrm{f}(\mathrm{x})=5 \mathrm{x}+\cos (2 \mathrm{x}+1)$ so $\mathrm{f}^{\prime}(\mathrm{x})=5-2 \sin (2 \mathrm{x}+1)$ which is defined for all values of $\mathrm{x}$. $\mathrm{f}^{\prime}(\mathrm{x})$ is always positve (why?) so $\mathrm{f}^{\prime}(\mathrm{x})$ is never equal to $0$. There are no endpoints. The function
$f(x)=5 x+\cos (2 x+1)$ is always increasing and has no critical numbers, no critical points, no local or global maximums or minimums.

13. $f(x)=e^{-(x-2)^{2}}$ so $f^{\prime}(x)=-2(x-2) e^{-(x-2)^{2}}$ which is defined for all values of $x$. $f^{\prime}(x)=0$ when $x=2$ so $x=2$ is a critical number. There are no endpoints. The only critical point is $(2,1)$ which is a local and global maximum. When the interval is the entire real number line, this function does not have a local or global minimum.

15. See Fig. 3.1P15

17. $f(x)=x^{2}-6 x+5$ on $[-2,5]$ so $f^{\prime}(x)=2 x-6$ which is defined for all values of $x$. $f^{\prime}(x)=0$ when $x=3$ so $x=3$ is a critical number. The endpoints are $x=-2$ and $x=5$ which are also critical numbers. The critical points are $(3,-4)$ which is the local and global minimum, $(-2,21)$ which is a local and global maximum, and $(5,0)$ which is a local maximum.

19. $f(x)=2-x^{3}$ on $[-2,1]$ so $f^{\prime}(x)=-3 x^{2}$ which is defined for all values of $x$. $f^{\prime}(x)=0$ when $x=0$ so $\mathrm{x}=0$ is a critical number. The endpoints are $\mathrm{x}=-2$ and $\mathrm{x}=1$ which are also critical numbers. The critical points are $(-2,10)$ which is a local and global maximum, $(0,2)$ which is not a local or global maximum or minimum, and $(1,1)$ which is a local and global minimum.

21. $f(x)=x^{3}-3 x+5$ on $[-2,1]$ so $f^{\prime}(x)=3 x^{2}-3=3(x-1)(x+1)$ which is defined for all values of $x$. $\mathrm{f}^{\prime}(\mathrm{x})=0$ when $\mathrm{x}=-1$ and $\mathrm{x}=+1$ so these are critical numbers. The endpoints $\mathrm{x}=-2$ and $\mathrm{x}=1$ are also critical numbers. The critical points are $(-2,3)$ which is a local and global minimum on $[-2,1]$, the point $(-1,7)$ which is a local and global maximum on $[-2,1]$, and the point $(1,3)$ which is a local and global minimum on $[-2,1]$.

23. $f(x)=x^{5}-5 x^{4}+5 x^{3}+7$ so $f^{\prime}(x)=5 x^{4}-20 x^{3}+15 x^{2}=5 x^{2}\left(x^{2}-4 x+3\right)=5 x^{2}(x-3)(x-1)$ which is defined for all values of $x$. $f^{\prime}(x)=0$ when $x=0$ and $x=1$ in the interval $[0,2]$ so each of these values is a critical number. The endpoints $x=0$ and $x=2$ are also critical numbers. The critical points are $(0,7)$ which is a local minimum, $(1,8)$ which is a local and global maximum, and $(2,-1)$ which is a local and global minimum. ($f^{\prime}(3)=0$ too, but $x=3$ is not in the interval $[0,2]$).

25. $f(x)=\frac{1}{x^{2}+1}$ so $f^{\prime}(x)=\frac{-2 x}{\left(x^{2}+1\right)^{2}}$ which is defined for all values of $x$. $f^{\prime}(x)=0$ when $x=0$ but $x=0$ is not in the interval $[1,3]$ so $x=0$ is a not a critcal number. The endpoints $x=1$ and $\mathrm{x}=3$ are critical numbers. The critical points are $(1,1 / 2)$ which is a local and global maximum, and $(3,1 / 10)$ which is a local and global minimum.

27. $\mathrm{A}(\mathrm{x})=4 \mathrm{x} \sqrt{1-\mathrm{x}^{2}}(0 < \mathrm{x} < 1)$
$A^{\prime}(x)=4\left[\frac{-x^{2}}{\sqrt{1-x^{2}}}+\sqrt{1-x^{2}}\right]=4 \frac{1-2 x^{2}}{\sqrt{1-x^{2}}} \begin{cases} > 0 & \text { if } 0 < x < 1 / \sqrt{2} \\ < 0 & \text { if } 1 / \sqrt{2} < x < 1\end{cases}$

A maximum is attained when $x=1 / \sqrt{2}: A(1 / \sqrt{2})=4 \frac{1}{\sqrt{2}} \sqrt{1-\frac{1}{2}}=4 \frac{1}{\sqrt{2}} \frac{1}{\sqrt{2}}=2$.

29. $\mathrm{V}=\mathrm{x}(8-2 \mathrm{x})^{2}$ for $0 < \mathrm{x} < 4$

$\begin{aligned}&\mathrm{V}^{\prime}=\mathrm{x}(2)(8-2 \mathrm{x})(-2)+(8-2 \mathrm{x})^{2}=(8-2 \mathrm{x})(-4 \mathrm{x}+8-2 \mathrm{x})=(8-2 \mathrm{x})(8-6 \mathrm{x})=4(4-\mathrm{x})(4-3 \mathrm{x}) \text { so } \\&\mathrm{V}^{\prime} \quad \begin{cases} < 0 & \text { if } 4 / 3 < \mathrm{x} \\ > 0 & \text { if } 0 < \mathrm{x} < 4 / 3\end{cases}\end{aligned}$

$\mathrm{V}(4 / 3)=\frac{4}{3}\left(8-\frac{8}{3}\right)^{2}=\frac{4}{3}\left(\frac{16}{3}\right)^{2}=\frac{1024}{27} \approx 37.926$ cubic units is the largest volume.

Smallest volume is $0$ which occurs when $x = 0$ and $x = 4$.

31. (a) $4$. The endpoints and two values of $x$ for which $f^{\prime}(x)=0$.
(b) $2$. The endpoints.
(c) At most $n+1$. The $2$ endpoints and the $n-1$ interior points $x$ for which $f^{\prime}(x)=0$. At least $2$. The $2$ endpoints.

33. (a) local minimum at $(1,5)$
(b) no extrema at $(1,5)$
(c) local maximum at $(1,5)$
(d) no extrema at $(1,5)$

35. (a) $0,2,6,8,11,12$
(b) $0,6,11$
(c) $2,8,12$

37. If f does not attain a maximum on $[a, b]$ or $f$ does not attain a mimimum on $[a, b]$, then $f$ must have a discontinuity on $[\mathrm{a}, \mathrm{b}]$.

39 (a) yes, $-1$
(b) no
(c) yes, $-1$
(d) no
(e) yes, $1-\pi$

41. (a) yes, $0$
(b) yes, $0$
(c) yes, $0$
(d) yes, $0$
(e) yes, $0$

43. (a) $\mathrm{S}(\mathrm{x})$ is minimum when $\mathrm{x} \approx 8$.
(b) $S(x)$ is maximum when $x=2$.