Using the second derivative f''(x) = 36x^2-12, how do you find the intervals in which f is concave up and the intervals in which f is concave down?

May 18, 2015

Unlike the first derivative test where it explains how a function is increasing or decreasing, we are now determining the concavity of a function, or how a line curves up or down. By taking the second derivative and setting up an interval. we can answer where the function $f \left(x\right)$ concaves up or down.

Given $f ' ' \left(x\right) = 36 {x}^{2} - 12$, our first step is to find the inflection points. They are points where the function's concavity shifts from down to up or vice versa, and results in no curvature instantly at that point (a line). $f \left(x\right) = {x}^{3}$ is a good example:
graph{x^3 [-5, 5, -5, 5]}
Notice that the concavity goes from down to up starting at the origin. This means that for ${x}^{3}$ $f ' ' \left(x\right)$ must be zero at $x = 0$.

So, we can say that $f ' ' \left(x\right) = 0$ for $f ' ' \left(x\right) = 36 {x}^{2} - 12$ to find the inflection points:

$0 = 36 {x}^{2} - 12 \implies 0 = 12 \left(3 {x}^{2} - 1\right)$
$0 = 12 \left(x \sqrt{3} + 1\right) \left(x \sqrt{3} - 1\right)$
$x = \pm \frac{\sqrt{3}}{3}$

Now, we should test some values around the inflection points to see if $f ' ' \left(x\right)$ is positive or negative at certain intervals:

For $x = 0$, $f ' ' \left(x\right) = - 12$; $f \left(x\right)$ concaves down for $x : \left(- \frac{\sqrt{3}}{3} , \frac{\sqrt{3}}{3}\right)$.
For $x = - 1$, $f ' ' \left(x\right) = + 24$; $f \left(x\right)$ concaves up for $x : \left(- \infty , - \frac{\sqrt{3}}{3}\right)$.
For $x = 1$, $f ' ' \left(x\right) = + 24$; $f \left(x\right)$ concaves up for $x : \left(\frac{\sqrt{3}}{3} , \infty\right)$.

Note that a line cannot curve on the inflection points. The parentheses () indicate the greater than ($>$) or less than ($<$) but not equal to $x$.

Here is a number line representation of the answer:

May 26, 2015

$f$ is concave up if $f ' '$ is positive ($f ' ' \left(x\right) > 0$) and

$f$ is concave down if $f ' '$ is negative ($f ' ' \left(x\right) < 0$)

We are given: $f ' ' \left(x\right) = 36 {x}^{2} - 12$

Use your method of choice for solving the inequalities:

$36 {x}^{2} - 12 > 0$ and $36 {x}^{2} - 12 < 0$.

I like finding partition number (zeros) and testing the resulting intervals.

For his problem:
$f ' ' \left(x\right)$ is never undefined.

$f ' ' \left(x\right) = 36 {x}^{2} - 12 = 12 \left(3 {x}^{2} - 1\right) = 0$ at $x = \pm \frac{1}{\sqrt{3}} = \pm \frac{\sqrt{3}}{3}$

On $\left(- \infty , - \frac{\sqrt{3}}{3}\right)$, we get $f ' '$ is positive, so $f$ is concave up.

On $\left(- \frac{\sqrt{3}}{3} , \frac{\sqrt{3}}{3}\right)$, we get $f ' '$ is negative, so $f$ is concave down.

On $\left(\frac{\sqrt{3}}{3} , \infty\right)$, we get $f ' '$ is positive, so $f$ is concave up.

Alternative for this $f ' '$

The graph of $f ' ' \left(x\right) = 36 {x}^{2} - 12$ is a parabola that opens upward. Between the zeros, $\pm \frac{\sqrt{3}}{3}$, we have $f ' '$ is negative and outside the zeros it is positive.