# How do you determine whether the function f^('')(x) = 36x^2-12 is concave up or concave down and its intervals?

Aug 3, 2015

You can use the second derivative test.

#### Explanation:

You can determine the intervals on which your function is concave up and concave down by using the second derivative test.

ALl you have to do is examine the behavior of the second derivative around inflexion points, which are points for which the second derivative is equal to zero.

${f}^{' '} = 36 {x}^{2} - 12$

Make ${f}^{' '} = 0$ and see how many inflexion points you have

$36 {x}^{2} - 12 = 0$

$12 \left(3 {x}^{2} - 1\right) = 0$

${x}^{2} = \frac{1}{3}$

Take the square root of both sides to get

$\sqrt{{x}^{2}} = \sqrt{\frac{1}{3}} \implies x = \pm \frac{\sqrt{3}}{3}$

You have two inflexion points, $x = - \frac{\sqrt{3}}{3}$ and $x = \frac{\sqrt{3}}{3}$, so you're going to look at three intervals

• $\left(- \infty , - \frac{\sqrt{3}}{3}\right)$

SInce you're dealing with a square number, the sign of ${f}^{' '}$ will always be positive for values that belong to this interval.

This means that $f \left(x\right)$ is concave up on this interval.

• $\left(- \frac{\sqrt{3}}{3} , \frac{\sqrt{3}}{3}\right)$

This time, the expression $\left(3 {x}^{2} - 1\right)$ will be negative for values that belong to this interval. This means that $f \left(x\right)$ will be concave down on this interval.

• $\left(\frac{\sqrt{3}}{3} , + \infty\right)$

Once again, the square of $x$ will ensure that the expression $\left(3 {x}^{2} - 1\right)$ will always be positive on this interval.

So, your function will be concave up on $\left(- \infty , - \frac{\sqrt{3}}{3}\right) \cup \left(\frac{\sqrt{3}}{3} , + \infty\right)$ and concave down on $\left(- \frac{\sqrt{3}}{3} , \frac{\sqrt{3}}{3}\right)$.

The graph of $f \left(x\right)$ will have two inflexion points at $\left(- \frac{\sqrt{3}}{3} , f \left(- \frac{\sqrt{3}}{3}\right)\right)$ and $\left(\frac{\sqrt{3}}{3} , f \left(\frac{\sqrt{3}}{3}\right)\right)$.