Mistakes when finding inflection points: not checking candidates

- [Instructor] 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. Alright, let's just follow her work. So here, she's trying to take the first derivative. So 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'd be three times four which would be 12 times x minus two to the second power times the derivative of x minus two which is just one which is exactly what she has here, 12 times x minus two 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 wanna make that equal to zero. This is only going to be true when x is equal to two. So step two is looking good. So step three, Olga says f has an inflection point at x equals two. So she's basing this just on the fact that the second derivative is zero when x is equal to two. She's basing this just on the fact that f prime prime of two is equal to zero. Now, I have a problem with this because the fact that your second derivative is zero at x equals two, that makes two a nice candidate to check out but 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 and speaking in the language of the second derivative, it means that the second derivative changes signs as we go from below x equals two to above x equals two 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 two and let's think about the interval where we go from two to positive infinity. If you want, you could have some test values. You could think about the sign, 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. So you could take a test value. Let's say one is in this interval and let's say three is in this interval and you could say one minus two squared is going to be, let's see, that's negative one squared which is one and then you're just going to, this is just going to be 12 so this is going to be positive and if you tried three, three minus two squared is one times 12. Well, that's also going to be positive and so you're going to be concave upwards. At least in these test values, it looks like on either side of two that the sign of the second derivative is positive on either side of the two and 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 two and this value right over here since we're, even if x minus two 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 two, the sign of our second derivative is positive which means that we're going to be concave upwards and 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 sides. So once again, this is incorrect. We actually don't have an inflection point at x equals two because our second derivative does not switch signs as we cross x equals two which means our concavity does not change.