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

##### 1 Answer

#### Answer:

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**.

In your case, you have

Make

Take the square root of both sides to get

You have *two inflexion points*, **three intervals**

#(-oo,-sqrt(3)/3)#

SInce you're dealing with a square number, the sign of **positive** for values that belong to this interval.

This means that **concave up** on this interval.

#(-sqrt(3)/3, sqrt(3)/3)#

This time, the expression **negative ** for values that belong to this interval. This means that **concave down** on this interval.

#(sqrt(3)/3, +oo)#

Once again, the square of **positive** on this interval.

So, your function will be *concave up* on *concave down* on

The graph of **two inflexion points** at