# What are the points of inflection, if any, of f(x) = x^4/12 - 2x^2 + 15 ?

Jul 27, 2017

The function $f \left(x\right) = {x}^{4} / 12 - 2 {x}^{2} + 15$ has two inflection points in ${x}_{1} = - 2$ and ${x}_{2} = 2$.

#### Explanation:

A necessary condition for the curve $y = f \left(x\right)$ to have an inflection point for $x = \overline{x}$ is that:

$f ' ' \left(\overline{x}\right) = 0$

Evaluate the second derivative of the function:

$f \left(x\right) = {x}^{4} / 12 - 2 {x}^{2} + 15$

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

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

From the equation:

${x}^{2} - 4 = 0$

${x}^{2} = 4$

$x = \pm 2$

we get that the function may have an inflection point in ${x}_{1} = - 2$ and ${x}_{2} = 2$.

Now consider the inequality:

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

As $f ' ' \left(x\right)$ is a second degree polynomial with positive leading coefficient, we have that its value is negative in the interval between the roots and positive outside.

$f ' ' \left(x\right)$ then changes sign both around ${x}_{1}$ and ${x}_{2}$, so these are actually inflection points.

Jul 27, 2017

There are two non-stationary points of inflection which occur at $\left(\pm 2 , \frac{25}{3}\right)$

#### Explanation:

We have:

$f \left(x\right) = {x}^{4} / 12 - 2 {x}^{2} + 15$

We would normally look for critical points, that is coordinates where $f ' \left(x\right) = 0$, however the question does not require this so it will be skipped.

The first derivative is then:

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

So the second derivative is then:

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

We look for inflection points , which are coordinates where the second derivative vanishes:

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

A point of inflection is graded as a stationary point of inflection of the first derivative vanished at the point, otherwise as a non-stationary point of inflection

So when $x = - 2$, we have:

$f \left(- 2\right) = \frac{16}{12} - 8 + 15 = \frac{25}{3}$
$f ' \left(- 2\right) = - \frac{8}{3} + 8 = \frac{16}{3}$

And when $x = 2$:

$f ' \left(2\right) = \frac{16}{12} - 8 + 15 = \frac{25}{3}$
$f ' \left(2\right) = \frac{8}{3} - 8 = \frac{16}{3}$

Hence, There are two non-stationary points of inflection which occur at $\left(\pm 2 , \frac{25}{3}\right)$

It can be interesting to see the graphs of the function compared with the first and second derivatives:

The graph of the function $y = f \left(x\right)$

graph{x^4/12-2x^2+15 [-6, 6, -10, 18]}

The graph of the function $y = f ' \left(x\right)$. The zeros coincide with critical points of the curve $y = f \left(x\right)$ ie the turning points of $y = f \left(x\right)$

graph{x^3/3-4x [-6, 6, -10, 18]}

The graph of the function $y = f ' ' \left(x\right)$. The zeros coincide with critical points of the curve $y = f ' \left(x\right)$ ie the turning points of $y = f ' \left(x\right)$ or the points of inflection of $y = f \left(x\right)$

graph{x^2-4 [-6, 6, -10, 18]}