Chapter 5.3, Problem 6E

To determine

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

Answer to Problem 6E

There is no local extremum and no absolute extremum.

Explanation of Solution

Given information :

The given function is $y=\{\begin{array}{l}3-x^2,x<0\\ x^2+1,x\ge 0\end{array}$ .

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 $f\left(x\right)=\{\begin{array}{l}3-x^2,x<0\\ x^2+1,x\ge 0\end{array}$ .

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

  $f\text{'}\left(x\right)=-2x,x<0$

And

  $f\text{'}\left(x\right)=2x,x\ge 0$

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

  $\begin{array}{l}-2x=0\\ x=0\end{array}$

And

  $\begin{array}{l}2x=0\\ x=0\end{array}$

Since, $f\text{'}\left(x\right)$ is always positive, therefore, there is no extreme values at $x=0$ .

From the first derivative test $f\text{'}$ does not change its sign. Therefore $f\left(x\right)$ does not have extreme values.

Also, since function is continuous for $-\infty <x<\infty$ . Therefore, there is no absolute extremum.

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

  

Hence,

There is no local extremum and no absolute extremum.