Chapter 5.3, Problem 3E

To determine

The local extreme values of the function, and identify any absolute extrema and support your answer graphically.

Answer to Problem 3E

The local minimum value is $-1$ at $x=-1,\text{ 1}$ and the local maximum value is 1 at $x=0$ and the absolute minimum is $-1$ at $x=-1,\text{ 1}$ .

Explanation of Solution

Given information :

The given function is $y=2x^4-4x^2+1$ .

Calculation:

For the continuous function $f\left(x\right)$ .

At the critical point $c$ :

1. If $f$ changes sign from positive to negative at $c$ ( $f\text{'}>c$ for $x<c$ and $f\text{'}<0$ for $x>c$ ), then $f$ has a local maximum value at $c$ .
2. if $f$ changes sign from negative to positive at $c$ ( $f\text{'}<0$ for $x<c$ and $f\text{'}>c$ for $x>c$ ), then $f$ has a local minimum value at $c$ .
3. if $f$ does not changes its sign at $c$ ( $f\text{'}$ has same sign on both sides of $c$ ), then $f$ has no local extreme value at $c$ .

Now let $f\left(x\right)=2x^4-4x^2+1$ .

Find the first derivative of $f\left(x\right)$ :

  $f\text{'}\left(x\right)=8x^3-8x$

Find the critical value by setting $f\text{'}\left(x\right)=0$ :

  $\begin{array}{l}8x^3-8x=0\\ 8x\left(x^2-1\right)=0\\ x\left(x-1\right)\left(x+1\right)=0\\ x=0\text{ or }x-1=0\text{ or }x+1=0\\ x=0\text{ or }x=1\text{ or }x=-1\end{array}$

The critical point partition the x-axis into four intervals as shown below:

  

From the first derivative test the sign of $f\text{'}$ is negative of $x<-1$ , positive for $-1<x<0$ negative for $0<x<1$ and again positive for $x>2$ .

Therefore, there is local minimum at $x=-1$ and at $x=1$ , the local maximum is at $x=0$ .

Now, the local minimum values are :

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

and

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

And, the local maximum value is

  $\begin{array}{l}f\left(0\right)=2{\left(0\right)}^4-4{\left(0\right)}^2+1\\ f\left(0\right)=0-0+1\\ f\left(0\right)=1\end{array}$

Since, from the sign chart the function for $x<-1$ and $x>1$ is increasing, therefore, the only time it goes as low as $-1$ is at $x=-1,\text{ 1}$ .

The graph of the function to support the answer is drawn below:

  

Hence,

The local minimum value is $-1$ at $x=-1,\text{ 1}$ and the local maximum value is 1 at $x=0$ and the absolute minimum is $-1$ at $x=-1,\text{ 1}$ .