Mistakes when finding inflection points: not checking candidates | AP Calculus AB | Khan Academy

Olga was asked to find where f of x is equal to x minus two to the fourth power has inflection points. This is her solution. So we look at her solution, and then they ask us: Is Olga's work correct? If not, what's her mistake?

So pause this video and see if you can figure this out.

All right, let's just follow her work. So here she's trying to take the first derivative. You would apply the chain rule: it would be four times x minus two to the third power times the derivative of x minus two, which is just one. So this checks out.

Then you take the derivative of this: it would be 3 times 4, which would be 12 times x minus 2 to the second power times the derivative of x minus 2, which is just 1. This is exactly what she has here: 12 times x minus 2 to the second power. That checks out. So step one's looking good for Olga.

Step two: the solution of the second derivative equaling zero is x equals two. That looks right. The second derivative is 12 times x minus two squared, and we want to make that equal to zero. This is only going to be true when x is equal to two. So step two is looking good.

Step three: Olga says f has an inflection point at x equals two. She's basing this just on the fact that the second derivative is 0 when x is equal to 2. Now, I have a problem with this because the fact that your second derivative is zero at x equals two makes two a nice candidate to check out. However, you can't immediately say that we have an inflection point there.

Remember, an inflection point is where we go from being concave upwards to concave downwards, or concave downwards to concave upwards. Speaking in the language of the second derivative, it means that the second derivative changes signs as we go from below x equals 2 to above x equals 2. But we have to test that, because it's not necessarily always the case.

So let's actually test it. Let's think about some intervals. Intervals? So let's think about the interval when we go from negative infinity to 2, and let's think about the interval where we go from 2 to positive infinity. If you want, you could have some test values; you could think about the sign of our second derivative, and then based on that, you could think about concavity—concavity of f.

So let's think about what's happening. You could take a test value. Let's say 1 is in this interval, and let's say 3 is in this interval. You could say 1 minus 2 squared is going to be, let's see, that's negative 1 squared, which is 1, and then you're just going to—this is just going to be 12. So this is going to be positive.

If you tried 3, 3 minus 2 squared is 1 times 12. Well, that's also going to be positive. So you're going to be concave upwards, at least at these test values. It looks like on either side of 2 that the sign of the second derivative is positive on either side of 2.

You might say, well, maybe I just need to find closer values. But if you inspect the second derivative here, you can see that this is never going to be negative. In fact, for any value other than x equals 2, this value right over here, since we're even if x minus 2 is negative, you're squaring it, which will make this entire thing positive, and then multiplying it times a positive value.

So for any value other than x equals 2, the sign of our second derivative is positive, which means that we're going to be concave upwards.

So we actually don't have an inflection point at x equals two because we are not switching signs as we go from values less than x equals two to values greater than x equals two. Our second derivative is not switching signs.

So once again, this is incorrect. We actually don't have an inflection point at x equals 2 because our second derivative does not switch signs as we cross x equals 2, which means our concavity does not change.