Chapter 4, Problem 1RE

To determine

To find: The local maximum and minimum and the absolute maximum and absolute minimum of the function $f\left(x\right)=x^3-9x^2+24x-2$ on the interval [0, 5].

Answer to Problem 1RE

The local maximum occurs at $x=2$.

The local minimum occurs at $x=4$.

The absolute maxima occurs at $x=2,5$.

The absolute minimum is $x=0$.

Explanation of Solution

Given:

The function is, $f\left(x\right)=x^3-9x^2+24x-2$ and the interval is, $\left[0,5\right]$.

Calculation:

Obtain the first derivative of the given function.

$f\text{'}\left(x\right)=3x^2-18x+24$

Set $f^{\prime }\left(x\right)=0$ and obtain the critical numbers.

$\begin{array}{l}3x^2-18x+24=0\\ 3\left(x-4\right)\left(x-2\right)=0\\ x=2,4\end{array}$

The critical numbers are 2 and 4 which lies on the given interval $\left[0,5\right]$.

Apply the extreme values of the given interval and the critical numbers in $f\left(x\right)$.

Substitute $x=0$ in $f\left(x\right)$,

$\begin{array}{c}f\left(0\right)=0^3-9{\left(0\right)}^2+24\left(0\right)-2\\ =-2\end{array}$

Substitute $x=2$ in $f\left(x\right)$,

$\begin{array}{c}f\left(2\right)=2^3-9{\left(2\right)}^2+24\left(2\right)-2\\ =8-9\left(4\right)+48-2\\ =18\end{array}$

Substitute $x=4$ in $f\left(x\right)$,

$\begin{array}{c}f\left(4\right)=4^3-9{\left(4\right)}^2+24\left(4\right)-2\\ =64-9\left(16\right)+96-2\\ =14\end{array}$

Substitute $x=5$ in $f\left(x\right)$,

$\begin{array}{c}f\left(5\right)=5^3-9{\left(5\right)}^2+24\left(5\right)-2\\ =125-9\left(25\right)+120-2\\ =18\end{array}$

Since the largest functional value is the absolute maximum and the smallest functional value is the absolute minimum, the absolute maximum of $f\left(x\right)$ is 18 and the absolute minimum of $f\left(x\right)$ is −2.

Therefore, the absolute maxima of $f\left(x\right)$ occurs at $x=2,5$ and the absolute minimum of $f\left(x\right)$ is occurs at $x=0$.

To find the local maximum and minimum obtain the second derivative of the function.

$\begin{array}{c}f\text{'}\text{'}\left(x\right)=\frac{d}{dx}\left(3x^2-18x+24\right)\\ =6x-18\end{array}$

Substitute the critical numbers in the second derivative and obtain the local maximum and minimum as follows.

Substitute $x=2$ in $f\text{'}\text{'}\left(x\right)$,

$\begin{array}{c}f\text{'}\text{'}\left(2\right)=6\left(2\right)-18\\ =12-18\\ =-6<0\end{array}$

Substitute $x=4$ in $f\text{'}\text{'}\left(x\right)$,

$\begin{array}{c}f\text{'}\text{'}\left(4\right)=6\left(4\right)-18\\ =24-18\\ =6>0\end{array}$

Thus, the local maximum occurs at $x=2$ and the local minimum occurs at $x=4$.