# 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?

##### 2 Answers

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

Given **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).

graph{x^3 [-5, 5, -5, 5]}

Notice that the concavity goes from down to up starting at the origin. This means that for

So, we can say that

Now, we should test some values around the inflection points to see if

For

For

For

Note that a line cannot curve on the inflection points. The parentheses () indicate the greater than (*not* equal to

Here is a number line representation of the answer:

We are given:

Use your method of choice for solving the inequalities:

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

For his problem:

On

On

On

**Alternative for this #f''#**

The graph of