(This is a continuation of Local Maximums and Minimums. It is recommended that you review the first and second derivative tests before going on.)
Inflection points are where the function changes concavity. Since concave up corresponds to a positive second derivative and concave down corresponds to a negative second derivative, then when the function changes from concave up to concave down (or vise versa) the second derivative must equal zero at that point. So the second derivative must equal zero to be an inflection point. But don't get excited yet. You have to make sure that the concavity actually changes at that point.
Example 1 with f(x) = x3.
Let's do an example to see what really happens. Given f(x) = x3, find the inflection point(s). (Might as well find any local maximum and local minimums as well.)
Start with getting the first derivative:
f '(x) = 3x2.
Then the second derivative is:
f "(x) = 6x.
Now set the second derivative equal to zero and solve for "x" to find possible inflection points.
6x = 0
x = 0.
We can see that if there is an inflection point it has to be at x = 0. But how do we know for sure if x = 0 is an inflection point? We have to make sure that the concavity actually changes. To do this pick a number on either side of x = 0 and check what the concavity is at those locations. Let's use x = -1 and x = 1 to check. At x = -1, the second derivative gives:
f "(-1) = -6
and the function is concave down at x = -1. If we check x = 1 we get:
f "(1) = 6
which means the function is concave up at x = 1.
Thus we can see that the function has different concavities on either side of x =0 and the inflection point is at x=0. Note the inflection point is not necessarily where the function crosses the x-axis but is where the concavity actually changes.
Let's now go back and find the local maximums and local minimums of this function. Start by finding the critical points.
f '(x) = 3x2
x = 0
We only have one critical point, x = 0. Is it a local max or a local min? Let's try using the second derivative test.
f "(x) = 6x
f "(0) = 6(0)
f "(0) = 0.
Well that's unfortunate because that means the function is neither concave up nor concave down. We still don't know if it is a local max or a local min. I guess we'll have to try another technique. Let's try the first derivative test.
Try using x=-1 and x=1 for numbers on either side of our critical point x=0. Plug them into the first derivative.
f '(-1) = 3(-1)2
f '(-1) = 3.
f '(1) = 3(1)2
f '(1) = 3.
Since the derivative is positive in either side of the critical point, the function is increasing on both side of the critical point and there is no local maximum or local minimum.
Example 2 with f(x) = x4.
Let's look at f(x) = x4. Set the derivative equal to zero to find the critical point(s).
f(x) = x4
f '(x) = 4x3 = 0
x3 = 0
x = 0
The only critical point is at x = 0. Let's try using the second derivative to test the concavity to see if it is a local maximum or a local minimum.
F "(x) = 12x2
f "(0) = 12(0)2 = 0
Since the second derivative is zero, the function is neither concave up nor concave down at x = 0. It could be still be a local maximum or a local minimum and it even could be an inflection point.
Let's test to see if it is an inflection point. We need to verify that the concavity is different on either side of x = 0. Let's test x = -1 and x = 1 in the second derivative.
f "(-1) = 12(-1)2 = 12
f "(1) = 12(1)2 = 12
Since the second derivative is positive on either side of x = 0, then the concavity is up on both sides and x = 0 is not an inflection point (the concavity does not change). Well it could still be a local maximum or a local minimum so let's use the first derivative test to find out.
f '(-1) = 4(-1)3 = -4
f '(1) = 4(1)3 = 4
Since the function goes from decreasing to increasing on either side of x = 0, we can see that x = 0 is a local minimum.
Even though f(x) = x4 appears to be concave up every where, it is momentarily "flat" at x = 0 since the second derivative is zero at x = 0.