What are the points of inflection, if any, of f(x)=2x^4-e^(8x?

Apr 24, 2018

See below

Explanation:

First step is finding the second derivative of the function

$f \left(x\right) = 2 {x}^{4} - {e}^{8 x}$

$f ' \left(x\right) = 8 {x}^{3} - 8 {e}^{8 x}$

$f ' ' \left(x\right) = 24 {x}^{2} - 64 {e}^{8 x}$

Then we must find a value of x where:

$f ' ' \left(x\right) = 0$

(I used a calculator to solve this)

$x = - 0.3706965$

So at the given $x$-value, the second derivative is 0. However, in order for it to be a point of inflection, there must be a sign change around this $x$ value.

Hence we can plug values into the function and see what happens:

$f \left(- 1\right) = 24 - 64 {e}^{- 8}$ definetly positive as $64 {e}^{- 8}$ is very small.

$f \left(1\right) = 24 - 64 {e}^{8}$ definetly negative as $64 {e}^{8}$ is very big.

So there is a sign change around $x = - 0.3706965$, so it is therefore an inflection point.