# How would you determine where the function f(x) is concave up and concave down for #f(x)=2x^3-3x^2-12x+1#?

##### 2 Answers

First, I would find the vertexes. Then, the inflection point.

The vertexes indicate where the slope of your function change, while the inflection points determine when a function changes from concave to convex (and vice-versa).

In order to find the vertexes (also named "points of maximum and minimum"), we must equal the first derivative of the function to zero, while to find the inflection points we must equal the second derivative to zero.

Using Bhaskara:

For

and

So, your vertexes are

Now, let's see the behaviour of your function relating to concavity.

For

Thus, your inflection point is

About concavity up or down: we just have one more thing to know about these vertexes, which is the way the function is taking after the vertex - "reading" it from left to right.

This is done by substituting the values we found for

Check the graph:

graph{2x^3-3x^2-12x+1 [-30.15, 32.28, -19.86, 11.34]}

To investigate concavity, we need to look at the sign of

In general a function can change signs by either crossing the

In the case, there are no discontinuities, so the only place **might** change signs is at:

Now we check the sign of

(This was not asked, but

**Note**

I prefer the general approach to finding the sign of

There are other methods for linear and quadratic

For example, in this problem