# What are the points of inflection of f(x)=4 / (2x^2?

Sep 8, 2016

No points of inflection.

#### Explanation:

We have: $f \left(x\right) = \frac{4}{2 {x}^{2}}$

First, let's evaluate the second derivative of this function using the "quotient rule":

$\implies f ' \left(x\right) = \frac{\left(2 {x}^{2} \cdot 0\right) - \left(4 \cdot 4 x\right)}{{\left(2 {x}^{2}\right)}^{2}}$

$\implies f ' \left(x\right) = - \frac{16 x}{4 {x}^{4}}$

$\implies f ' \left(x\right) = - \frac{4}{{x}^{3}}$

$\implies f ' ' \left(x\right) = \frac{\left({x}^{3} \cdot 0\right) - \left(- 4 \cdot 3 {x}^{2}\right)}{{\left({x}^{3}\right)}^{2}}$

$\implies f ' ' \left(x\right) = \frac{12 {x}^{2}}{{x}^{6}}$

$\implies f ' ' \left(x\right) = \frac{12}{{x}^{4}}$

Then, to determine the points of inflection, we need to set the second derivative equal to zero, and then solve for $x$:

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

$\implies \frac{12}{{x}^{4}} = 0$

$\implies 12 \ne 0$

$\therefore$ no real solutions

Therefore, there are no points of inflection for $f \left(x\right)$.