Chapter 4.3, Problem 4QQ

a.

To determine

To find: The relative minimum and relative maximum.

Answer to Problem 4QQ

The relative minimum at $x=1$ and $x=4$

The relative maximum $x=\pm 2$ .

Explanation of Solution

Given:

The function $f(x)=3\ln (x^2+2)-2x$

  $[-2,4]$ .

Calculation:

A point in the interior of the domain of a function $f$ at which $f\text{'}=0$ or $f\text{'}$ does not exist is a critical point of $f$ .

To find critical point,

  $\begin{array}{l}f\text{'}(x)=0\\ \frac{d}{dx}3\ln (x^2+2)-2x=0\\ 3\left(\frac{2x}{x^2+2}\right)-2=0\\ x=1\\ x=2\end{array}$

With keeping the critical values and end points in mind obtain the following sign chart:

Intervals	$-2<x<1$	$1<x<2$	$2<x<4$
Sign of $y\text{'}$	-	+	-
Behaviour of $y$	Decreasing	Increasing	Decreasing

The chart shows that the function $f$ has relative minimum at $x=1$ and $x=4$

Also, $f$ has relative maximum at $x=\pm 2$ .

b.

To determine

To find: the point of deflection.

Answer to Problem 4QQ

The point of deflection is $x=\pm \sqrt{2}$

Explanation of Solution

Given:

The function $f(x)=3\ln (x^2+2)-2x$

  $[-2,4]$ .

Calculation:

The second derivative of $f$ is

  $\begin{array}{l}f\text{'}\text{'}(x)=\frac{d}{dx}\left(\frac{6x}{x^2+2}-2\right)\\ f\text{'}\text{'}(x)=\frac{6}{x^2+2}-\frac{12x^2}{{\left(x^2+2\right)}^2}\end{array}$

To find the point of deflection

  $\begin{array}{l}f\text{'}\text{'}(x)=0\\ \frac{6}{x^2+2}-\frac{12x^2}{{\left(x^2+2\right)}^2}=0\\ x=\pm \sqrt{2}\end{array}$

c.

To determine

To find: the absolute maximum

Answer to Problem 4QQ

Absolute maxima at $x=-2$ and $f(-2)=3\ln 6+4$

Explanation of Solution

Given:

The function $f(x)=3\ln (x^2+2)-2x$

  $[-2,4]$ .

Calculation:

At $x=-2$ it is found that $f\text{'}\text{'}(x)<0$ so the absolute maximum is at $x=-2$ and the absolute maximum value of $f(x)$ is

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