# How do you find the inflection points for y= e^(2x) - e^x ?

May 6, 2015

I changed the format of your question, because I think you want to find the inflection points for $y = {e}^{2 x} - {e}^{x}$
(Not for $y = {e}^{2} x - {e}^{x}$ which has none.)

$y = {e}^{2 x} - {e}^{x}$

$y ' = 2 {e}^{2 x} - {e}^{x}$

$y ' ' = 4 {e}^{2 x} - {e}^{x}$ . The partition numbers for $y ' '$ are found by:

$4 {e}^{2 x} - {e}^{x} = {e}^{x} \left(4 {e}^{x} - 1\right) = 0$ .

Since ${e}^{x}$ is never $0$, the only partition number is the solution to:

$\left(4 {e}^{x} - 1\right) = 0$

${e}^{x} = \frac{1}{4}$

$x = \ln \left(\frac{1}{4}\right) = - \ln 4 = - 2 \ln 2$

$y ' ' = {e}^{x} \left(4 {e}^{x} - 1\right)$ .

The first factor is always positive and the second is negative for $x < \ln \left(\frac{1}{4}\right)$ and positive for $x > \ln \left(\frac{1}{4}\right)$, so the sign of $y ' '$ does change at the partition number.

The only inflection point is at the point where $x = \ln \left(\frac{1}{4}\right)$

At ${e}^{x} = \frac{1}{4}$, it is easy to calculate $y$ because ${e}^{2 x} = {\left(\frac{1}{4}\right)}^{2} = \frac{1}{16}$, so

$y = \frac{1}{16} - \frac{1}{4} = - \frac{3}{16}$
(Or, if you prefer, rewrite $y = {e}^{x} \left({e}^{x} + 1\right)$ and use that to find $y$.)

The only inflection point is $\left(- \ln 4 , - \frac{3}{16}\right)$