Chapter 5.3, Problem 35E

To determine

To find : The local extrema of the function by using second derivative test.

Answer to Problem 35E

The local maximum is at $x=-2$ and local minimum is at $x=0$ .

Explanation of Solution

Given information :

The given function is $y=x^3+3x^2-2$ .

Calculation :

The second derivative test for local extrema is:

If $f\text{'}\left(c\right)=0$ and $f\text{'}\text{'}\left(c\right)<0$ , then $f$ has a local maximum at $x=c$ . And

If $f\text{'}\left(c\right)=0$ and $f\text{'}\text{'}\left(c\right)|>0$ , then $f$ has a local minimum at $x=c$ .

Let

  $f\left(x\right)=x^3+3x^2-2$ .

Differentiate $f\left(x\right)$ :

  $\begin{array}{l}f\left(x\right)=x^3+3x^2-2\\ f\text{'}\left(x\right)=3x^2+6x\end{array}$

For critical point set $f\text{'}\left(x\right)=0$ and solve for x:

  $\begin{array}{l}3x^2+6x=0\\ 3x\left(x+2\right)=0\\ x\left(x+2\right)=0\\ x=0\text{ or }x+2=0\\ x=0\text{ or }x=-2\end{array}$

Now differentiate $f\text{'}\left(x\right)$ again:

  $\begin{array}{l}f\text{'}\left(x\right)=3x^2+6x\\ f\text{'}\text{'}\left(x\right)=6x+6\end{array}$

Now,

  $\begin{array}{l}f\text{'}\text{'}\left(0\right)=6\left(0\right)+6\\ f\text{'}\text{'}\left(0\right)=6>0\text{ ( local minimum)}\end{array}$

And

  $\begin{array}{l}f\text{'}\text{'}\left(-2\right)=6\left(-2\right)+6\\ f\text{'}\text{'}\left(-2\right)=-6<0\text{ ( local maximum)}\end{array}$

Hence,

The local maximum is at $x=-2$ and local minimum is at $x=0$ .