# What are the points of inflection of f(x)= e^(2x) - e^x ?

Nov 21, 2016

$x = - \ln 4$
$x \approx - 1.3863$

#### Explanation:

Points of inflection occur when the second derivative is equal to zero. Find this by first differentiating $f \left(x\right)$ to get $f ' \left(x\right)$, then differentiating $f ' \left(x\right)$ to get $f ' ' \left(x\right)$.

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

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

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

Set $f ' ' \left(x\right)$ equal to zero to find possible points of inflection:
$0 = 4 {e}^{2 x} - {e}^{x}$
${e}^{x} = 4 {e}^{2 x}$
Rewrite as a natural log:
$x = \ln \left(4 {e}^{2 x}\right)$
$x = \ln 4 + \ln \left({e}^{2 x}\right)$
$x = \ln 4 + 2 x \ln e$
$x = \ln 4 + 2 x$
$- x = \ln 4$

$x = - \ln 4$
$x \approx - 1.3863$
Check if this is a point of inflection by making sure $f ' ' \left(x\right)$ is positive on one side of the x value, and negative on the other (make a sign chart):
$\text{- "x=-ln4" +}$

Therefore, $x = - \ln 4$ is the point of inflection of $f \left(x\right) = {e}^{2 x} - {e}^{x}$