Chapter 11.3, Problem 64E

To determine

To calculate:

The points present on a graph of the following functions where the horizontal tangent line is observed:

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

Answer to Problem 64E

The point at that the function and tangent drawn its horizontal are $\left(0,0\right),\left(1,e^{-1}\right)$

Explanation of Solution

Given information:

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

Calculation:

Consider this function:

  $f\left(x\right)=x{e}^{-x}$

The derivative or the slope of the function $f\text{'}\left(x\right)$ is $f\text{'}\left(x\right)=-x{e}^{-x}+e^{-x}$

The slope of tangent drawn on this function $f\left(x\right)=x{e}^{-x}$ must be horizontal.

The slope will be zero.

  $\begin{array}{l}-x{e}^{-x}+e^{-x}=0\\ e^{-x}\left(-x+1\right)=0\\ x=0,1\end{array}$

Now, $x=0,1$ has to be on function because at point the function and tangent will touch to each other. So that $x=0,-2$ should satisfy this function.

When, $x=0$

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

When, $x=1$

  $\begin{array}{l}f\left(1\right)={\left(1\right)}^2{e}^{-1}\\ f\left(0\right)=e^{-1}\end{array}$

The point at that the tangent and function drawn it is horizontal are $\left(0,0\right),\left(1,e^{-1}\right)$

  

Graph of function:

Therefore, the point $\left(0,0\right),\left(1,e^{-1}\right)$ lies on the graph of the function $f\left(0\right)=x{e}^{-x}$ , it has verified that at the point $\left(0,0\right),\left(1,e^{-1}\right)$ the tangents drawn on the function $f\left(0\right)=x{e}^{-x}$ is horizontal.