Chapter 5.3, Problem 38E

To determine

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

Answer to Problem 38E

The local maximum is at $x=1$ .

Explanation of Solution

Given information :

The given function is $y=x{e}^{-x}$ .

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

  $f\left(x\right)=x{e}^{-x}$ .

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

  $\begin{array}{l}f\left(x\right)=x{e}^{-x}\\ f\text{'}\left(x\right)=e^{-x}-x{e}^{-x}\\ f\text{'}\left(x\right)=e^{-x}\left(1-x\right)\end{array}$

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

  $\begin{array}{l}{e}^{-x}\left(1-x\right)=0\\ 1-x=0\text{ (Since, }e\text{ is never zero)}\\ x=1\end{array}$

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

  $\begin{array}{l}f\text{'}\left(x\right)=e^{-x}\left(1-x\right)\\ f\text{'}\text{'}\left(x\right)=\left(-e^{-x}\right)\left(1-x\right)+e^{-x}\left(-1\right)\\ f\text{'}\text{'}\left(x\right)=e^{-x}\left(x-1\right)-e^{-x}\\ f\text{'}\text{'}\left(x\right)=e^{-x}\left[\left(x-1\right)-1\right]\\ f\text{'}\text{'}\left(x\right)=e^{-x}\left(x-2\right)\end{array}$

Now,

  $\begin{array}{l}f\text{'}\text{'}\left(1\right)=e^{-1}\left(1-2\right)\\ f\text{'}\text{'}\left(1\right)=-\frac{1}{e}<0\text{ ( local maximum)}\end{array}$

Hence,

The local maximum is at $x=1$ .