Chapter 4.3, Problem 8E

(a)

To determine

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

Explanation of Solution

Given: The function is $f\left(x\right)=4x^3+3x^2-6x+1$

Find the first derivative of the given function.

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

Construct the table as shown below.

Interval	$x+1$	$2x-1$	$f^{\prime }\left(x\right)$	$f$
$x<-1$	$-$	$-$	$+$	Increasing on $\left(-\infty ,-1\right)$
$-1<x<0.5$	$+$	$-$	$-$	Decreasing on $\left(-1,0.5\right)$
$x>0.5$	$+$	$+$	$+$	Increasing on $\left(0.5,\infty \right)$

Table (1)

Therefore, the function increases in the interval $\left(-\infty ,-1\right)\cup \left(0.5,\infty \right)$ and decrease on $\left(-1,0.5\right)$

(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 $0.5$ .

  $\begin{array}{c}f\left(0.5\right)=4{\left(0.5\right)}^3+3{\left(0.5\right)}^2-6\left(0.5\right)+1\\ =0.5+0.75-3+1\\ =-0.75\end{array}$

So, the local minima is $f\left(0.5\right)=-0.75$ .

The value of $f^{\prime }\left(x\right)$ changes from positive to negative at $-1$ .

  $\begin{array}{c}f\left(-1\right)=4{\left(-1\right)}^3+3{\left(-1\right)}^2-6\left(-1\right)+1\\ =-4+3+6+1\\ =6\end{array}$

So, the local maxima is $f\left(-1\right)=6$ .

(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.

  $\begin{array}{c}{f}^{″}\left(x\right)=24x+6\\ =6\left(4x+1\right)\end{array}$

Consider the table shown below.

Interval	$f^{″}\left(x\right)=6\left(4x+1\right)$	Concavity
$\left(-\infty ,-0.25\right)$	$-$	Downward
$\left(-0.25,\infty \right)$	$+$	Upward

Table (2)

The function changes it sign at $-0.25$ . So, the point of inflection is $-0.5$

  $\begin{array}{c}f\left(-0.25\right)=4{\left(-0.25\right)}^3+3{\left(-0.25\right)}^2-6\left(-0.25\right)+1\\ =-0.0625+.01875+2.5\\ =2.625\end{array}$

Therefore, the function is concave upward at $\left(-0.25,\infty \right)$ , concave downward at $\left(-\infty ,-0.25\right)$ and the point of inflection is $\left(-0.25,2.625\right)$ .