# Could I get some help determining the point of inflection of a function?

1. Feb 25, 2009

### meeklobraca

1. The problem statement, all variables and given/known data

Determine the point of inflection on the graph y = x^4 - 4x^3

2. Relevant equations

3. The attempt at a solution

The second derivative of this function is 12x^2 - 24x

I dont know how to show my work to get a point of inflection of (0,0)

How would I go about doing that?

2. Feb 25, 2009

### Unassuming

Set the second derivative equal to zero and solve. You should get two values and then check a trial point on both sides of each of those in order to make sure that the concavity changes.

3. Feb 25, 2009

### meeklobraca

How do I do a trial to see the concavity changes?

I figured here my x values were x = 0 and x = 2. Do I plug these values into the second derivative to determine if the concavity changes?

4. Feb 26, 2009

### Staff: Mentor

Yes. At each of these points y'' should change from pos to neg or neg to pos.

5. Feb 26, 2009

### meeklobraca

I used the original function to check to see if it changes from positive to negative and vise versa, not y". When I used y" it told me x=2 is a point of inflection when I can see on the graph that it is not.

6. Feb 26, 2009

### meeklobraca

Oh I think I got it wrong. x=2 is a point of inflection. Could someone verify for me? I get (0,0) and (2,-16) as points of infection for this function.

7. Feb 26, 2009

### HallsofIvy

Yes, the second derivative is, as you said, $f"(x)= 12x^2- 24x= 12x(x- 2)$. If x< 0, both x and x-2 are negative so f" is positive. If 0< x< 2, x is positive while x- 2 is still negative so f" is negative. If 2< x, both x and x-2 are positive so f" is positive. f" changes sign at both 0 and 2 so 0 and 2 are inflection points.