Chapter 5.3, Problem 4QQ

a.

To determine

To find: the coordinate of each relative maximum point and each relative minimum point of f .

Answer to Problem 4QQ

Relative minimum at x = 1 and x = 4. Relative maximum at x = 2 and x = -2.

Explanation of Solution

Given information: Given function is $\mathrm{f}(\mathrm{x})=3\ln ({\mathrm{x}}^2+2)-2\mathrm{x}$ with domain [-2, 4].

Calculation:

  $\begin{array}{c}\mathrm{f}(\mathrm{x})=3\ln ({\mathrm{x}}^2+2)-2\mathrm{x}\\ \mathrm{f}\text{'}(\mathrm{x})=3.\frac{1}{({\mathrm{x}}^2+2)}.2\mathrm{x}-2=\frac{6\mathrm{x}}{({\mathrm{x}}^2+2)}-2\\ \mathrm{f}\text{'}(\mathrm{x})=0\Rightarrow \frac{6\mathrm{x}}{({\mathrm{x}}^2+2)}-2=0\\ \frac{6\mathrm{x}}{({\mathrm{x}}^2+2)}-\frac{2({\mathrm{x}}^2+2)}{({\mathrm{x}}^2+2)}=0\\ \frac{-2{\mathrm{x}}^2+6\mathrm{x}-4}{({\mathrm{x}}^2+2)}=0\\ -2{\mathrm{x}}^2+6\mathrm{x}-4=0\\ -2(\mathrm{x}-1)(\mathrm{x}-2)=0\Rightarrow \mathrm{x}=1.\mathrm{x}=2\end{array}$

Plug in a value between -2 and 1 to see if the function is increasing or decreasing on that interval.

  $\mathrm{f}\text{'}(0)=\frac{-2{(0)}^2+6(0)-4}{{(0)}^2+2}=-2<0$

Plug in a value between 1 and 2 to see if the function is increasing or decreasing on that interval

  $\mathrm{f}\text{'}(1.5)=\frac{-2{(1.5)}^2+6(1.5)-4}{{(1.5)}^2+2}\approx 0.118>0$

Plug in a value between 2 and 4 to see if the function is increasing or decreasing on that interval.

  $\mathrm{f}\text{'}(3)=\frac{-2{(3)}^2+6(3)-4}{{(3)}^2+2}=-\frac{4}{11}<0$

There is a relative max at x = -2 because it is decreasing from -2 to 1 and at an endpoint, if the points following are smaller, it is a maximum. There is a min at x = 1 because the interval goes from decreasing to increasing. There is max at x = 2 because interval goes from increasing to decreasing. At x = 4 there is a minimum because the interval coming into x = 4 is decreasing.

Therefore, relative minimum at x = 1 and x = 4. Relative maximum at x = 2 and x = -2

b.

To determine

To find: the x −coordinate of each pint of inflection of the graph of f .

Answer to Problem 4QQ

  $\mathrm{x}=\pm \sqrt{2}$ .

Explanation of Solution

Given information: Given function is $\mathrm{f}(\mathrm{x})=3\ln ({\mathrm{x}}^2+2)-2\mathrm{x}$ with domain [-2, 4].

Calculation:

  $\begin{array}{c}\mathrm{f}(\mathrm{x})=3\ln ({\mathrm{x}}^2+2)-2\mathrm{x}\\ \mathrm{f}\text{'}(\mathrm{x})=3.\frac{1}{({\mathrm{x}}^2+2)}.2\mathrm{x}-2=\frac{6\mathrm{x}}{({\mathrm{x}}^2+2)}-2\\ \mathrm{u}=6\mathrm{x},\mathrm{u}\text{'}=6,\mathrm{v}={\mathrm{x}}^2+2,\mathrm{v}\text{'}=2\mathrm{x}\\ \mathrm{f}\text{'}\text{'}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{d}\mathrm{x}}\left(\frac{6\mathrm{x}}{({\mathrm{x}}^2+2)}-2\right)=\frac{6({\mathrm{x}}^2+2)-12{\mathrm{x}}^2}{{({\mathrm{x}}^2+2)}^2}=\frac{12-6{\mathrm{x}}^2}{{({\mathrm{x}}^2+2)}^2}\\ \mathrm{f}\text{'}\text{'}(0)=0\Rightarrow \frac{12-6{\mathrm{x}}^2}{{({\mathrm{x}}^2+2)}^2}=0\\ 12-6{\mathrm{x}}^2=0\\ {\mathrm{x}}^2=2\\ \mathrm{x}=\pm \sqrt{2}\end{array}$

Plug in a value between 2 and 4 to see if the function is increasing or decreasing on that interval.

  $\mathrm{f}\text{'}(3)=\frac{-2{(3)}^2+6(3)-4}{{(3)}^2+2}=-\frac{4}{11}<0$

c.

To determine

To find: the absolute maximum value of f(x).

Answer to Problem 4QQ

The absolute maximum value is $\mathrm{f}(-2)=3\ln (6)+4$ .

Explanation of Solution

Given information: Given function is $\mathrm{f}(\mathrm{x})=3\ln ({\mathrm{x}}^2+2)-2\mathrm{x}$ with domain [-2, 4].

Calculation:

Two possible max values occur at x = 2 and x = -2.

  $\begin{array}{c}\mathrm{f}(\mathrm{x})=3\ln ({\mathrm{x}}^2+2)-2\mathrm{x}\\ $ f(-2)=3\ln ({(-2)}^2+2)-2(-2)=3\ln (6)+4$$ f(2)=3\ln ({(2)}^2+2)-2(2)=3\ln (6)-4$\Rightarrow \mathrm{f}(-2)>\mathrm{f}(2)\end{array}$

Since f (-2) is bigger than f (2), it is the absolute maximum value.

Therefore, the absolute maximum value is $\mathrm{f}(-2)=3\ln (6)+4$ .

Plug in a value between 2 and 4 to see if the function is increasing or decreasing on that interval.

  $\mathrm{f}\text{'}(3)=\frac{-2{(3)}^2+6(3)-4}{{(3)}^2+2}=-\frac{4}{11}<0$