# What are the inflection points of f(x)?

## $f \left(x\right) = 1.5 {x}^{5} - 5 {x}^{4} - 20 {x}^{3} + 120 {x}^{2} - 2 x - 6$

Mar 29, 2018

$\left(- 2 , 510\right)$ is the only point of inflection.

#### Explanation:

We must first find the second derivative.

$f ' \left(x\right) = 7.5 {x}^{4} - 20 {x}^{3} - 60 {x}^{2} + 240 x - 2$
$f ' ' \left(x\right) = 30 {x}^{3} - 60 {x}^{2} - 120 x + 240$

Inflection points occur when $f ' ' \left(x\right) = 0$.

$0 = 30 {x}^{3} - 60 {x}^{2} - 120 x + 240$

Thankfully we can readily solve this by factoring.

$0 = 30 {x}^{2} \left(x - 2\right) - 120 \left(x - 2\right)$

$0 = \left(30 {x}^{2} - 120\right) \left(x - 2\right)$

$x = 2 \mathmr{and} {x}^{2} = 4$

$x = \pm 2$

But before declaring these, we must test the sign of the second derivative between these points. An inflection point only occurs if the sign changes.

Consider the following table:

As you can see, the second derivative does equal $0$ at $x = 2$, but it does not goes from positive to negative. Instead it stays positive. Therefore, we must reject $x = 2$.

However, $x = - 2$ does have a sign change, therefore it's acceptable.

Hopefully this helps!