# Find the inflection point of the function e^(-x/2) ?

Sep 22, 2017

There are no inflection points.

#### Explanation:

The inflection point will be where the second derivative equals $0$. By definition, it's when the graph goes from concave up to concave down, or vice versa.

Calling the function $f \left(x\right)$, we have:

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

$f ' \left(x\right) = - \frac{1}{2} {e}^{- \frac{x}{2}}$

$f ' ' \left(x\right) = \frac{1}{4} {e}^{- \frac{x}{2}}$

If we set this to $0$, we get:

$0 = \frac{1}{4} {e}^{- \frac{x}{2}}$

Which doesn't have any solutions because ${e}^{x} \ne 0$. So there are no inflection points.

Hopefully this helps!