Practice Problems

1. Label all of the local maximums and minimums of the function in Fig. 13. Also label all of the critical points.

Fig. 13

In problems 3-13, find all of the critical points and local maximums and minimums of each function.

3. $f(x)=x^{2}+8 x+7$

5. $f(x)=\sin (x)$

7. $f(x)=(x-1)^{2}(x-3)$

9. $f(x)=2 x^{3}-96 x+42$

11. $f(x)=5 x+\cos (2 x+1)$

13. $f(x)=e^{-(x-2)^{2}}$

15. Sketch the graph of a continuous function $\mathrm{f}$ so that

(a) $f(1)=3, f^{\prime}(1)=0$, and the point $(1,3)$ is a relative maximum of $f$.
(b) $\mathrm{f}(2)=1$, $\mathrm{f}^{\prime}(2)=0$, and the point $(2,1)$ is a relative minimum of $\mathrm{f}$.
(c) $\mathrm{f}(3)=5$, $\mathrm{f}$ is not differentiable at $3$, and the point $(3,5)$ is a relative maximum of $\mathrm{f}$.
(d) $\mathrm{f}(4)=7$, $\mathrm{f}$ is not differentiable at $4$, and the point $(4,7)$ is a relative minimum of $\mathrm{f}$.
(e) $f(5)=4, f^{\prime}(5)=0$, and the point $(5,4)$ is not a relative minimum or maximum of $f$.
(f) $\mathrm{f}(6)=3$, $\mathrm{f}$ is not differentiable at $6$, and the point $(6,3)$ is not a relative minimum or maximum of $\mathrm{f}$.

In problems 17-25, find all critical points and local extremes of each function on the given intervals.

17. $f(x)=x^{2}-6 x+5$ on $[-2,5]$.

19. $f(x)=2-x^{3}$ on $[-2,1]$.

21. $f(x)=x^{3}-3 x+5$ on $[-2,1]$.

23. $f(x)=x^{5}-5 x^{4}+5 x^{3}+7$ on $[0,2]$.

25. $\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{x}^{2}+1}$ on $[1,3]$.

Fig. 15

29. Find the value for $\mathrm{x}$ so the box in Fig. 17 has the largest possible volume? The smallest volume?

Fig. 17

31. Suppose you are working with a polynomial of degree $3$ on a closed interval.
(a) What is the largest number of critical points the function can have on the interval?
(b) What is the smallest number of critical points it can have?
(c) What are the patterns for the most and fewest critical points a polynomial of degree $\mathrm{n}$ on a closed interval can have?

33. Suppose $f(1)=5$ and $f^{\prime}(1)=0$. What can we conclude about the point $(1,5)$ if
(a) $\mathrm{f}^{\prime}(\mathrm{x}) < 0$ for $\mathrm{x} < 1$, and $\mathrm{f}^{\prime}(\mathrm{x}) > 0$ for $\mathrm{x} > 1$?
(b) $\mathrm{f}^{\prime}(\mathrm{x}) < 0$ for $\mathrm{x} < 1$, and $\mathrm{f}^{\prime}(\mathrm{x}) < 0$ for $\mathrm{x} > 1$?
(c) $\mathrm{f}^{\prime}(\mathrm{x}) > 0$ for $\mathrm{x} < 1$, and $\mathrm{f}^{\prime}(\mathrm{x}) < 0$ for $\mathrm{x} > 1$?
(d) $\mathrm{f}^{\prime}(\mathrm{x}) > 0$ for $\mathrm{x} < 1$, and $\mathrm{f}^{\prime}(\mathrm{x}) > 0$ for $\mathrm{x} > 1$?

35. $\mathrm{f}$ is a continuous function, and Fig. 18 shows the graph of $\mathrm{f} '$
(a) Which values of $\mathrm{x}$ are critical points?
(b) At which values of $\mathrm{x}$ is $\mathrm{f}$ a local maximum?
(c) At which values of $\mathrm{x}$ is $\mathrm{f}$ a local minimum?

Fig. 18

37. State the contrapositive form of the Extreme Value Theorem.

39. Imagine the graph of $\mathrm{f}(\mathrm{x})=1-\mathrm{x}$. Does $\mathrm{f}$ have a minimum value for $\mathrm{x}$ in the interval $\mathrm{I}$?
(a) $\mathrm{I}=[0,2]$
(b) $I=[0,2)$
(c) $\mathrm{I}=(0,2]$
(d) $\mathrm{I}=(0,2)$
(e) $\mathrm{I}=(1, \pi]$

41. Imagine the graph of $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}$. Does $\mathrm{f}$ have a minimum value for $\mathrm{x}$ in the following intervals?
(a) $\mathrm{I}=[-2,3]$
(b) $\mathrm{I}=[-2,3)$
(c) $\mathrm{I}=(-2,3]$
(d) $\mathrm{I}=[-2,1)$
(e) $\mathrm{I}=(-2,1]$

43. Define $S(x)$ to be the slope of the line through the points $(0,0)$ and $(\mathrm{x}$, $\mathrm{f}(\mathrm{x}))$ in Fig. 21 .
(a) At what value of $\mathrm{x}$ is $\mathrm{S}(\mathrm{x})$ minimum?
(b) At what value of $\mathrm{x}$ is $\mathrm{S}(\mathrm{x})$ maximum?

Fig. 21

Source: Dale Hoffman, https://s3.amazonaws.com/saylordotorg-resources/wwwresources/site/wp-content/uploads/2012/12/MA005-4.1-Finding-Maximums-and-Minimums.pdf
This work is licensed under a Creative Commons Attribution 3.0 License.

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 $x=2$.