# Analyzing Exponential Functions

Now we’ll take a look at an example where we analyze an exponential function. The method is as follows:

Rule

### AnalyzingExponentialFunctions

1.
Find the zeros.
2.
Find the stationary points.
3.
Find the inflection points.

Example 1

Analyze the function $f\left(x\right)=2{x}^{2}\cdot {e}^{x}$

1.
Find the zeros by setting $f\left(x\right)=0$:
 $2{x}^{2}\cdot {e}^{x}=0$

The zero product property gives that $2{x}^{2}=0$ or ${e}^{x}=0$. However, ${e}^{x}$ is always positive, so $\begin{array}{llll}\hfill 2{x}^{2}& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill {x}^{2}& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill x& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

This gives a zero at the origin $\left(0,0\right)$.

2.
Find the maxima and minima by setting ${f}^{\prime }\left(x\right)=0$.

First, find the derivative of $f\left(x\right)=2{x}^{2}\cdot {e}^{x}$: $\begin{array}{llll}\hfill {f}^{\prime }\left(x\right)& =4x\cdot {e}^{x}+2{x}^{2}\cdot {e}^{x}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={e}^{x}\phantom{\rule{-0.17em}{0ex}}\left(4x+2{x}^{2}\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2x\cdot {e}^{x}\left(2+x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Then, find where ${f}^{\prime }\left(x\right)$ is equal to 0:

 $2x\cdot {e}^{x}\left(2+x\right)=0$

Again, ${e}^{x}$ is always positive, so $\begin{array}{llllllllllll}\hfill 2x& =0\phantom{\rule{2em}{0ex}}& \hfill ⇒& \phantom{\rule{2em}{0ex}}& \hfill x& =0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill 2+x& =0\phantom{\rule{2em}{0ex}}& \hfill ⇒& \phantom{\rule{2em}{0ex}}& \hfill x& =-2\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

You then need the corresponding $y$-values to find the point. You do this by inputting your $x$-values back into the main function $f\left(x\right)$:

$\begin{array}{llll}\hfill y& =f\left(0\right)=2\cdot {0}^{2}\cdot {e}^{0}=0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y& =f\left(-2\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2{\left(-2\right)}^{2}\cdot {e}^{-2}=8{e}^{-2}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =\frac{8}{{e}^{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill y& =f\left(0\right)=2\cdot {0}^{2}\cdot {e}^{0}=0\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y& =f\left(-2\right)=2{\left(-2\right)}^{2}\cdot {e}^{-2}=8{e}^{-2}=\frac{8}{{e}^{2}}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

You now need to determine which point is a maximum and which is a minimum. You do that by drawing a sign chart.

From this, you see that the maximum is $\phantom{\rule{-0.17em}{0ex}}\left(-2,\frac{8}{{e}^{2}}\right)$ and the minimum is $\left(0,0\right)$.
3.
Find the inflection points by setting ${f}^{″}\left(x\right)=0$.

First, you find the second derivative by differentiating ${f}^{\prime }\left(x\right)={e}^{x}\left(4x+2{x}^{2}\right)$:

$\begin{array}{llll}\hfill {f}^{″}\left(x\right)& ={e}^{x}\phantom{\rule{-0.17em}{0ex}}\left(4x+2{x}^{2}\right)+{e}^{x}\left(4+4x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={e}^{x}\phantom{\rule{-0.17em}{0ex}}\left(4x+2{x}^{2}+4+4x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={e}^{x}\phantom{\rule{-0.17em}{0ex}}\left(2{x}^{2}+8x+4\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill {f}^{″}\left(x\right)& ={e}^{x}\phantom{\rule{-0.17em}{0ex}}\left(4x+2{x}^{2}\right)+{e}^{x}\left(4+4x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={e}^{x}\phantom{\rule{-0.17em}{0ex}}\left(4x+2{x}^{2}+4+4x\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & ={e}^{x}\phantom{\rule{-0.17em}{0ex}}\left(2{x}^{2}+8x+4\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

Then, let ${f}^{″}\left(x\right)=0$ and solve the equation:
 ${e}^{x}\phantom{\rule{-0.17em}{0ex}}\left(2{x}^{2}+8x+4\right)=0$

As ${e}^{x}$ is always positive, you get

 $2{x}^{2}+8x+4=0$

You solve this using the quadratic formula and get the solutions $x\approx -0.6$ and $x\approx -3.4$. You find the corresponding $y$-values by putting your new $x$-values back into the main function $f\left(x\right)$. You then get: $\begin{array}{llll}\hfill y& =f\left(-3.4\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2\cdot {\left(-3.4\right)}^{2}\cdot {e}^{-3.4}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 0.772\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill y& =f\left(-0.6\right)\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2{\left(-0.6\right)}^{2}\cdot {e}^{-0.6}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =8{e}^{-0.6}\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \approx 0.395\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

which means that you have inflection points at $\left(-3.4,\phantom{\rule{0.17em}{0ex}}0.772\right)$ and $\left(-0.6,\phantom{\rule{0.17em}{0ex}}0.395\right)$.