Chapter 1, Problem 9CT

To determine

the open intervals on which the functions is increasing, decreasing or constant.

Answer to Problem 9CT

Increasing intervals: $\left(-2,0\right)$ , $\left(2,\infty \right)$

Decreasing intervals: $\left(-\infty ,-2\right)$ , $\left(0,2\right)$

Explanation of Solution

Given information :

The function is $h\left(x\right)=\frac{1}{4}{x}^4-2x^2$.

Calculation : the domain of this function is all real numbers, and its critical points occur at $-2,0,2$ . So first differentiate the function ,

  $\begin{array}{l}h\left(x\right)=\frac{1}{4}{x}^4-2x^2\\ h\text{'}\left(x\right)=\frac{4}{4}{x}^3-2·2x\text{ [differentiate function]}\\ h\text{'}\left(x\right)=x^3-4x\end{array}$

For the interval $\left(-\infty ,-2\right)$ ,

  $h\text{'}\left(x\right)<0$

So for this interval the function is decreasing.

For the interval $\left(-2,0\right)$

  $h\text{'}\left(x\right)>0$

So for this interval function is increasing.

For the interval $\left(0,2\right)$

  $h\text{'}\left(x\right)<0$

So for this interval function is decreasing.

For the interval $\left(2,\infty \right)$

  $h\text{'}\left(x\right)>0$

So for this interval function is increasing