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

Dec 24, 2015

Crap.

#### Explanation:

Was utter crap so forget I said anything.

Dec 24, 2015

There is an inflection point at $x = - 2 \ln \left(2\right)$

#### Explanation:

To find inflection points, we apply the second derivative test.

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

We apply the second derivative test by setting $f ' ' \left(x\right)$ equal to $0$.

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

One property of logarithms is that terms being multiplied in a single logarithm can be turned into a sum of logarithms for each term:

$\ln \left(4 {e}^{2 x}\right) = \ln \left({e}^{x}\right)$
$\ln \left(4\right) + \ln \left({e}^{2 x}\right) = \ln \left({e}^{x}\right)$
$\ln \left(4\right) + 2 x = x$
$x = - \ln \left(4\right)$
$x = - \ln \left({2}^{2}\right)$
$x = - 2 \ln \left(2\right) \approx - 1.3863 \ldots$

Although you usually don't see inflection points with exponentials, the fact that one is being subtracted from the other means that there is the possibility of them "affecting" the graph in ways that offers the possibility for an inflection point.

graph{e^(2x) - e^(x) [-4.278, 1.88, -1.63, 1.447]}

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

You can see that the portion of the line left of the point appears to be concave down, whereas the portion to the right changes and becomes concave up.