Chapter 4.3, Problem 7E

(a)

To determine

To find:The interval on which $f$ is increasing.

Explanation of Solution

Given: The function is $f\left(x\right)=2x^3+3x^2-36x$

Find the first derivative of the given function.

  $\begin{array}{c}{f}^{\prime }\left(x\right)=6x^2+6x-36\\ =6\left(x^2+x-6\right)\\ =6\left(x+3\right)\left(x-2\right)\end{array}$

Construct the table as shown below.

Interval	$x+3$	$x-2$	$f^{\prime }\left(x\right)$	$f$
$x<-3$	$-$	$-$	$+$	Increasing on $\left(-\infty ,-3\right)$
$-3<x<2$	$+$	$-$	$-$	Decreasing on $\left(-3,2\right)$
$x>2$	$+$	$+$	$+$	Increasing on $\left(2,\infty \right)$

Table (1)

Therefore, the function increases in the interval

  $\left(-\infty ,-3\right)\cup \left(2,\infty \right)$ and decrease on $\left(-3,2\right)$

(b)

To determine

To find :The local maxima and minima.

Explanation of Solution

Consider table (1).

The value of $f^{\prime }\left(x\right)$ changes from negative to positive at $2$ .

  $\begin{array}{c}f\left(2\right)=2{\left(2\right)}^3+3{\left(2\right)}^2-36\left(2\right)\\ =16+12-72\\ =-44\end{array}$

So, the local minima is $f\left(2\right)=-44$ .

The value of $f^{\prime }\left(x\right)$ changes from positive to negative at $-3$ .

  $\begin{array}{c}f\left(-3\right)=2{\left(-3\right)}^3+3{\left(-3\right)}^2-36\left(-3\right)\\ =-54+27+108\\ =81\end{array}$

So, the local maxima is $f\left(-3\right)=81$ .

(c)

To determine

To find :The interval ofconcavity and points of inflection

Explanation of Solution

The graph is concave up when $f^{″}\left(x\right)>0$ and concave downward when $f^{″}\left(x\right)<0$ .

The derivative of $f^{\prime }\left(x\right)$ is positive when $f^{\prime }\left(x\right)$ is increasing.

The derivative of $f^{\prime }\left(x\right)$ is negative when $f^{\prime }\left(x\right)$ is decreasing

Calculate the second derivative.

  $\begin{array}{c}{f}^{″}\left(x\right)=12x+6\\ =6\left(2x+1\right)\end{array}$

Consider the table shown below.

Interval	$f^{″}\left(x\right)=6\left(2x+1\right)$	Concavity
$\left(-\infty ,-0.5\right)$	$-$	Downward
$\left(-0.5,\infty \right)$	$+$	Upward

Table (2)

The function changes it sign at $-0.5$ . So, the point of inflection is $-0.5$

  $\begin{array}{c}f\left(-0.5\right)=2{\left(-0.5\right)}^3+3{\left(-0.5\right)}^2-36\left(-0.5\right)\\ =-0.25+0.75+18\\ =18.5\end{array}$

Therefore, the function is concave upward at $\left(-0.5,\infty \right)$ , concave downward at $\left(-\infty ,-0.5\right)$ and the point of inflection is $\left(-0.5,18.5\right)$ .