Chapter 5.3, Problem 2E

To determine

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

Answer to Problem 2E

The local minimum value is $-3$ at $x=0$ and the local maximum value is $5$ at $x=2$ and there is no absolute extrema.

Explanation of Solution

Given information :

The given function is $y=-2x^3+6x^2-3$ .

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 $f\left(x\right)=-2x^3+6x^2-3$ .

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

  $f\text{'}\left(x\right)=-6x^2+12x$

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

  $\begin{array}{l}-6x^2+12x=0\\ -6x\left(x-2\right)=0\\ x\left(x-2\right)=0\\ x=0\text{ or }x-2=0\\ x=0\text{ or }x=2\end{array}$

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

  

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

Therefore, there is local minimum at $x=0$ and local maximum at $x<2$

Now, the local minimum value is

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

And, the local maximum value is

  $\begin{array}{l}f\left(2\right)=-2{\left(2\right)}^3+6{\left(2\right)}^2-3\\ f\left(2\right)=-2\left(8\right)+6\left(4\right)-3\\ f\left(2\right)=-16+24-3\\ f\left(2\right)=5\end{array}$

Since, the function has a range $\left(-\infty ,\infty \right)$ , therefore, there is no absolute extrema.

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

  

Hence,

The local minimum value is $-3$ at $x=0$ and the local maximum value is $5$ at $x=2$ and there is no absolute extrema.