Chapter 12.3, Problem 68E

To determine

Verify that the tangent line is horizontal.

Answer to Problem 68E

Tangents drawn on the function $f(x)=x{e}^{-x}$ are horizontal.

Explanation of Solution

Given information:

Use the function and its derivative to determine any points on the graph of $f$ at which the tangent line is horizontal. Use a graphing utility to verify the results.

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

Calculation:

Consider the function, $f(x)=x{e}^{-x}$

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

The slope of the tangent drawn on the function $f(x)=x{e}^{-x}$ should be horizontal.

So, the slope will be $0$

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

  $x=$

  $0$ , $1$

Now, $x=$

  $0$ , $1$ should satisfy the function.

When, $x=$

  $0$

  $f(0)=0\times e^{-0}$

  $=0$

When, $x=$

  $1$

  $f(1)=e^{-1}$

So, the points at which the function $f(x)=x{e}^{-x}$ and tangents drawn it is horiontal are $\boxed{\left(0,0\right)}$ and $\boxed{\left(1,e^{-1}\right)}$ .

Now to check the above point by calculator, proceed as follows.

Press button $"=Y"$ and then enter the expression in the window as given below.

  

Now, press the $"WINDOW"$ button to set the range.

  

Now, press the $"GRAPH"$ button to plot the graph as for different point .

For point $\left(0,0\right)$

  

For point $\left(1,e^{-1}\right)$

  

Hence, it is verified that the tangents drawn on the function $f(x)=x{e}^{-x}$ are horizontal.