# What are the points of inflection, if any, of f(x)= 35x^4-36x^2 + 5x ?

Feb 3, 2016

$x = \pm \sqrt{\frac{6}{35}}$

#### Explanation:

A point of inflection occurs when the concavity of a function shifts. This is also when the sign of the second derivative shifts--that is, the second derivative switches from positive to negative, or vice versa.

First, find the second derivative through the power rule:

$f \left(x\right) = 35 {x}^{4} - 36 {x}^{2} + 5 x$

$f ' \left(x\right) = 140 {x}^{3} - 72 x$

$f ' ' \left(x\right) = 420 {x}^{2} - 72$

The sign of the second derivative could shift when the second derivative equals $0$.

$420 {x}^{2} - 72 = 0$

$420 {x}^{2} = 72$

${x}^{2} = \frac{6}{35}$

$x = \pm \sqrt{\frac{6}{35}}$

There are two possible points of inflection at $x = - \sqrt{\frac{6}{35}}$ and $x = \sqrt{\frac{6}{35}}$. However, they are not guaranteed to be the spot where the sign shifts, so we should plug in values surrounding these points.

We can test the intervals surrounding the two possible points of inflection to see if the sign of the second derivative does actually shift.

When $x < - \sqrt{\frac{6}{35}}$:

$f ' ' \left(- 1\right) = 420 {\left(- 1\right)}^{2} - 72 = 348$

When $- \sqrt{\frac{6}{35}} < x < \sqrt{\frac{6}{35}}$:

$f ' ' \left(0\right) = 420 {\left(0\right)}^{2} - 72 = - 72$

When $x > \sqrt{\frac{6}{35}}$:

$f ' ' \left(1\right) = 420 {\left(1\right)}^{2} - 72 = 348$

Notice that at $x = - \sqrt{\frac{6}{35}}$, the sign of the second derivative switches from positive to negative. Also, at $x = \sqrt{\frac{6}{35}}$, the sign switches from negative to positive. Since the sign switches in both of these places, these are both points of inflection.