Chapter 4.3, Problem 9E

(a)

To determine

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

Explanation of Solution

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

Find the first derivative of the given function.

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

The function increases in the interval $\left(-1,0\right)\cup \left(1,\infty \right)$ and decreases in the interval $\left(-\infty ,-1\right)\cup \left(0,1\right)$ .

Therefore, the function increases in the interval $\left(-1,0\right)\cup \left(1,\infty \right)$ and decreases in the interval $\left(-\infty ,-1\right)\cup \left(0,1\right)$ .

(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 $1$ and $-1$

So, the local minima is $x=-1,1$ .

The value of $f^{\prime }\left(x\right)$ changes from positive to negative at $0$

So, the local maxima is $x=0$ .

(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.

  $\begin{array}{c}{f}^{″}\left(x\right)=12x^2-4\\ =4\left(3x^2-1\right)\end{array}$

Equate to zero.

  $\begin{array}{c}0=3x^2-1\\ x^2=\frac{1}{3}\\ x=\pm \sqrt{\frac{1}{3}}\\ =\pm \frac{\sqrt{3}}{3}\end{array}$

Therefore, the function is concave upward at $\left(-\infty ,-\sqrt{\frac{1}{3}}\right)\cup \left(\sqrt{\frac{1}{3}},\infty \right)$ , concave downward at $\left(-\sqrt{\frac{1}{3}},\sqrt{\frac{1}{3}}\right)$ and the point of inflection is $\pm \frac{\sqrt{3}}{3}$ .