Chapter 11.3, Problem 63E

To determine

To calculate:

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

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

Answer to Problem 63E

The point at that the function and the tangent drawn it is horizontal are $\left(0,0\right)\left(-2,4e^{-2}\right)$

Explanation of Solution

Given information:

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

Calculation:

Consider this function:

  $f\left(x\right)=x^2{e}^x$

The derivative or the slope of the function $f\text{'}\left(x\right)$ is $f\text{'}\left(x\right)=x^2{e}^x+2x{e}^x$

The slope of tangent drawn on this function $f\left(x\right)=x^2{e}^x$ should be horizontal.

The slope will be zero.

  $\begin{array}{l}{x}^2{e}^x+2x{e}^x=0\\ x^2{e}^x\left(x+2\right)=0\\ x=0,-2\end{array}$

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

When, $x=0$

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

When, $x=-2$

  $\begin{array}{l}f\left(-2\right)={\left(-2\right)}^2{e}^x\\ f\left(0\right)=4e^{-2}\end{array}$

The point at that the function and the tangent drawn it is horizontal are $\left(0,0\right)\left(-2,4e^{-2}\right)$

  

Graph of function:

Therefore, the point $\left(0,0\right)\left(-2,4e^{-2}\right)$ lies on the graph of the function $f\left(0\right)=x^2{e}^x$ , it has verified that at the point $\left(0,0\right)\left(-2,4e^{-2}\right)$ the tangents drawn on the function $f\left(0\right)=x^2{e}^x$ are horizontal.