Chapter 4, Problem 6RE

To determine

To find: The local maximum and minimum and the absolute maximum and absolute minimum of the function $f\left(x\right)=x^2{e}^{-x}$ on the interval [-1, 3].

Answer to Problem 6RE

The local maximum occurs at $x=2$.

The local minimum occurs at $x=0$.

The absolute maximum occurs at $x=-1$.

The absolute minimum is $x=0$.

Explanation of Solution

Given:

The function is, $f\left(x\right)=x^2{e}^{-x}$ and the interval is, [-1, 3].

Calculation:

Obtain the first derivative of the given function.

$f\text{'}\left(x\right)=2x{e}^{-x}-x^2{e}^{-x}$

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

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

The critical numbers are 0 and 2 which lies on the given interval $\left[-1,3\right]$.

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

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

$\begin{array}{c}f\left(0\right)=0^2{e}^{-0}\\ =0\end{array}$

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

$\begin{array}{c}f\left(2\right)=2^2{e}^{-2}\\ =\frac{4}{e^2}\\ \approx 0.54\end{array}$

Substitute $x=-1$ in $f\left(x\right)$,

$\begin{array}{c}f\left(-1\right)={\left(-1\right)}^2{e}^{-\left(-1\right)}\\ =e\\ \approx 2.72\end{array}$

Substitute $x=3$ in $f\left(x\right)$,

$\begin{array}{c}f\left(3\right)=3^2{e}^{-3}\\ =\frac{9}{e^3}\\ \approx 0.45\end{array}$

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 2.72 and the absolute minimum of $f\left(x\right)$ is 0.

Therefore, the absolute maximum of $f\left(x\right)$ occurs at $x=-1$ and the absolute minimum of $f\left(x\right)$ occurs at $x=0$.

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

$\begin{array}{c}f\text{'}\text{'}\left(x\right)=\frac{d}{dx}\left(2x{e}^{-x}-x^2{e}^{-x}\right)\\ =2e^{-x}-2x{e}^{-x}-2x{e}^{-x}+x^2{e}^{-x}\\ =e^{-x}\left(x^2-4x+2\right)\end{array}$

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

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

$\begin{array}{c}f\text{'}\text{'}\left(2\right)=e^{-2}\left(4-8+2\right)\\ =\frac{-2}{e^2}\\ =-0.45\end{array}$

Substitute $x=0$ in $f\text{'}\text{'}\left(x\right)$,

$\begin{array}{c}f\text{'}\text{'}\left(0\right)=e^0\left(0-0+2\right)\\ =2\end{array}$

Thus, the local maximum occurs at $x=2$ and the local minimum occurs at $x=0$.