# How do you identify a critical point of y = e^(x+2) – e^(2x) and determine whether it is a local maximum or a local minimum?

Mar 29, 2015

George I. has given a fine solution. I want to add that there's no need for technology to test the critical number.

$f \left(x\right) = {e}^{x + 2} - {e}^{2 x}$

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

${e}^{x + 2} - 2 {e}^{2 x} = 0$ iffi ${e}^{x + 2} = 2 {e}^{2 x}$ iffi $x + 2 = \ln \left(2\right) + 2 x$

So the critical points is $c = 2 - \ln 2$.

Now how big is this number?

$0 < \ln 2 < 1$
so $- 1 < - \ln 2 < 0$
and $2 - 1 < 2 - \ln 2 < 2$ (add 2 to all 3 parts)

$1 < 2 - \ln 2 < 2$

Use the first derivative test.

$0 < c$ and $f ' \left(0\right) = {e}^{2} - 2$ which is positive since ${e}^{2} > 2$. So $f$ is increasing left of $c$

$c < 2$ and $f ' \left(2\right) = {e}^{2} - 2 {e}^{2} = - {e}^{2}$ which is negative so $f$ is decreasing right of $c$.

$f \left(c\right) = f \left(2 - \ln 2\right)$ is a local maximum.

If you want to use the second derivative test:

$f ' ' \left(x\right) = {e}^{x + 2} - 4 {e}^{2 x}$

f''(2- ln 2)= e^((2- ln 2)+2)-4e^(2(2- ln 2)) =e^((2- ln 2)+2)-4e^(4- 2ln 2))

$f \left(c\right) = f \left(2 - \ln 2\right)$ is a local maximum.

$= {e}^{2} / {e}^{\ln 2} - 4 {e}^{4} / {e}^{\ln 4} = {e}^{2} / 2 - 4 {e}^{4} / 4 = - {e}^{2} / 2$ which is negative, so