Chapter 4.3, Problem 44E

(a)

To determine

To estimate the maximum and minimum values.

Explanation of Solution

Given:

The function is

  $f\left(x\right)=x^2{e}^{-x}$

Concept used:

The slope of the tangent to a curve $f\left(x\right)$ at any point a is given by the value of the first derivative of the slope at the point $x=a$ .

The tangent to be horizontal so the slope should be equal to 0

That is $\frac{dy}{dx}=0$ .

Calculation:

The function is

  $f\left(x\right)=x^2{e}^{-x}\mathrm{................................}\left(1\right)$

Differentiating equation (1) with respect to $x$

  $\begin{array}{l}f\left(x\right)=x^2{e}^{-x}\\ f^{\prime }\left(x\right)=2x{e}^{-x}-x^2{e}^{-x}\mathrm{.............................}\left(2\right)\end{array}$

  $f^{\prime }\left(x\right)$ is the maximum value

Draw the table

  $f\left(x\right)=x^2{e}^{-x}$

Test one point in each of the region formed by the graph

If the point satisfies the function then shade the entire region to denote that every point in the region satisfies the function

$x-axis$	$0$	$-2$	$-1$	$1.6$
$y-axis$	$0$	$0.5$	$0.36$	$0$

Draw the graph

  $f\left(x\right)=x^2{e}^{-x}$

  

(b)

To determine

To find: The exact value of $x$ at $f$ increases most rapidly .

Explanation of Solution

Given:

The function is

  $f\left(x\right)=x^2{e}^{-x}$

Concept used:

The slope of the tangent to a curve $f\left(x\right)$ at any point a is given by the value of the first derivative of the slope at the point $x=a$ .

The tangent to be horizontal so the slope should be equal to 0

That is $\frac{dy}{dx}=0$ .

Calculation:

The function is

  $f\left(x\right)=x^2{e}^{-x}\mathrm{................................}\left(1\right)$

Differentiating equation (1) with respect to $x$

  $\begin{array}{l}f\left(x\right)=x^2{e}^{-x}\\ f^{\prime }\left(x\right)=2x{e}^{-x}-x^2{e}^{-x}\mathrm{.............................}\left(2\right)\end{array}$

Again differentiating equation (2) with respect to $x$

  $\begin{array}{l}{f}^{″}\left(x\right)=2e^{-x}-2x{e}^{-x}-2x{e}^{-x}+x^2{e}^{-x}\\ f^{″}\left(x\right)=\left(2-4x+x^2\right)e^{-x}\\ f^{″}\left(x\right)=0\\ \left(2-4x+x^2\right)e^{-x}=0\\ \text{since}{e}^{-x}\ne 0\\ \left(2-4x+x^2\right)=0\\ x=2\pm \sqrt{2},x=2-\sqrt{2}\end{array}$

Therefore ,

The exact value is $x=2\pm \sqrt{2},x=2-\sqrt{2}$ .