# How do you find the inflection points for the function f(x)=e^(3x)+e^(-2x)?

Mar 16, 2018

There are no points of inflections.

#### Explanation:

First, calculate the first and the second derivatives

$f \left(x\right) = {e}^{3 x} + {e}^{- 2 x}$

The domain of $f \left(x\right)$ is $x \in \mathbb{R}$

$f ' \left(x\right) = 3 {e}^{3 x} - 2 {e}^{- 2 x}$

$f ' ' \left(x\right) = 9 {e}^{3 x} + 4 {e}^{- 2 x}$

The points of inflections are when $f ' ' \left(x\right) = 0$

$f ' ' \left(x\right) = 9 {e}^{3 x} + 4 {e}^{- 2 x} = 9 {e}^{3 x} + \frac{4}{e} ^ \left(2 x\right)$

$= \frac{9 {e}^{5 x} + 4}{e} ^ \left(2 x\right)$

$\forall x \in \mathbb{R} , f ' ' \left(x\right) > 0$

So there are no points of inflections

graph{e^(3x)+e^(-2x) [-6.52, 4.58, 0.71, 6.26]}