# How do you find points of inflection for (x) = 2x(x-4)^3?

May 14, 2015

There are two points of inflection : one when ${x}_{1} = 2$ and another when ${x}_{2} = 4$.

The points of inflection of a function are given by the study of the sign the second derivative. Those are the points where the concavity of a function changes.

Let's derive $f \left(x\right) = 2 x {\left(x - 4\right)}^{3}$ :

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

$f ' \left(x\right) = 2 {\left(x - 4\right)}^{3} + 6 x {\left(x - 4\right)}^{2} = {\left(x - 4\right)}^{2} \left(2 x - 8 + 6 x\right)$

$f ' \left(x\right) = 8 {\left(x - 4\right)}^{2} \left(x - 1\right)$

Now, let's derive $f ' \left(x\right)$ :

$f ' ' \left(x\right) = \left({\left(x - 4\right)}^{2}\right) ' \cdot 8 \cdot \left(x - 1\right) + \left(x - 1\right) ' \cdot 8 \cdot {\left(x - 4\right)}^{2}$

$f ' ' \left(x\right) = 16 \left(x - 4\right) \left(x - 1\right) + 8 {\left(x - 4\right)}^{2}$

$f ' ' \left(x\right) = \left(x - 4\right) \left(16 x - 16 + 8 x - 32\right)$

$f ' ' \left(x\right) = \left(x - 4\right) \left(24 x - 48\right) = 24 \left(x - 4\right) \left(x - 2\right)$

$f ' ' \left(x\right) = 0$, when ${x}_{1} = 2$ and ${x}_{2} = 4$.

Therefore, your function has two points of inflection : one when ${x}_{1} = 2$ and another when ${x}_{2} = 4$.

That's it!