Chapter 2.8, Problem 28E

a.

To determine

The first derivative and second derivative of the equation.

Explanation of Solution

Given information:

The equation is $f(x)=x^4-2x^2$

Calculations:

Here given the equation is $f(x)=x^4-2x^2$ .

The defination of derivative,is $f\text{'}(x)={\lim }_{h\to 0}\frac{f(x+h)-f(x)}{h}$

Now according to the defination of derivative,

  $\begin{array}{l}f\text{'}(x)={\lim }_{h\to 0}\frac{\left[{(x+h)}^4-2{(x+h)}^2\right]-\left(x^4-2x^2\right)}{h}\\ \text{Simplifying the equation we get,}\\ f\text{'}(x)=4x^3-4x\end{array}$

Now,The defination of second derivative,is $f\text{'}(x)={\lim }_{h\to 0}\frac{f\text{'}(x+h)-f\text{'}(x)}{h}$

Using the formula we get the second derivative of the equation,

  $\begin{array}{l}f\text{'}(x)=4x^3-4x\\ f\text{'}\text{'}(x)=12x^2-4\end{array}$

b.

To determine

The interval in which $f$ is increasing or decreasing.

Explanation of Solution

Given information:

The equation is $f(x)=x^4-2x^2$

Calculations:

Here using the equation we draw the graph of the equation.

  

Now from the graph it can be easily understood that, from $x=-\infty \text{ to }x=1$ and $x=0\text{ to }x=1$ $f$ is decreasing.

From the graph it can be easily understood that, for $x=-1\text{ to }x=0$ and $x=1\text{ to }x=\infty$ $f$ is increasing.

c.

To determine

The interval in which $f$ concave upward and downward.

Explanation of Solution

Given information:

The equation is $f(x)=x^4-2x^2$

Calculations:

Here using the equation we draw the graph of the equation.

  

Now from the graph it can be easily understood that, from $x=-\infty \text{ to }x=1$ and $x=1\text{ to }x=\infty$ $f$ concave upward

From the graph it can be easily understood that, for $x=-1\text{to }x=1$ $f$ concave downward.