Mistakes when finding inflection points: second derivative undefined | AP Calculus AB | Khan Academy

Robert was asked to find where ( g(x) ), which is equal to the cube root of ( x ), has inflection points. This is his solution, and then later we are asked if Robert's work is correct. If not, what's his mistake? So pause this video and try to figure it out on your own.

All right, now let's work through this together. So our original ( g(x) ) is equal to the cube root of ( x ), which is the same thing as ( x^{\frac{1}{3}} ).

In step one, it looks like Robert's trying to find the first and second derivative. The first derivative, we just do the power rule, so it will be ( \frac{1}{3} x^{-\frac{2}{3}} ). So this is looking good. For the second derivative, we take this and multiply it by ( \frac{1}{3} ), which would give us ( -\frac{2}{9} x^{-\frac{5}{3}} ). So that looks right.

Then it looks like Robert's trying to rewrite it. We have ( -\frac{2}{9} ) still, but then he recognized that this is the same thing as ( x^{\frac{5}{3}} ) in the denominator. ( x^{\frac{5}{3}} ) is also the same thing as the cube root of ( x^{5} ). So this is all looking good.

Step one looks good, and then in step two, it looks like he's trying to find the solution or he's trying to find ( x ) values where the second derivative is equal to zero. It is indeed true that this second derivative has no solution; you can never make this second derivative equal to zero. In order to be zero, the numerator would have to be zero and, well, ( 2 ) is never going to be equal to zero. So this is correct.

Then in step three, he says ( g ) doesn't have any inflection points. Now this is a little bit suspect. In many cases, our inflection point is a situation where our second derivative is equal to zero, and even then, we don't know it's an inflection point. It would be a candidate inflection point; we would have to confirm that our second derivative crosses signs or switches signs as we cross that ( x ) value.

But here we can't find a situation where our second derivative is equal to zero. However, we have to remind ourselves that other candidate inflection points are where our second derivative is undefined. So he can't make this statement without seeing where our second derivative could be undefined.

For example, he could say that ( g'' ) is undefined when ( x ) is equal to zero. ( x^{0} ) raised to the fifth is ( 0 ), so the cube root of that is going to be ( 0 ), but then you're dividing by ( 0 ). So ( g'' ) is undefined when ( x ) is equal to ( 0 ).

Therefore, we could say a candidate inflection point exists when ( x = 0 ). So then we would want to test it, and we could set up a traditional table that you might have seen before where we have our intervals. We could have test values in our intervals. We have to be careful with those; make sure that they are indicative.

Then we would say the sign of our second derivative ( g'' ) and then we would have our concavity of ( g ). In order for ( x = 0 ) to be an inflection point, we would have to switch signs as our second derivative would have to switch signs as we cross ( x = 0 ). This would mean our concavity of ( g ) switches signs as we cross ( x = 0 ).

So let's do values less than zero (( -\infty ) to ( 0 )) and then values greater than zero (( 0 ) to ( \infty )). I could do test values; let's say I'll use ( -1 ) and ( 1 ). You have to be careful when you use these; you have to make sure that we are close enough that nothing unusual happens between these test values up until we get to that candidate inflection point.

Now what's the sign of our second derivative when ( x ) is equal to ( -1 )? When ( x = -1 ), let's see: ( (-1)^{5} ) is ( -1 ). The cube root of ( -1 ) is ( -1 ). So we're going to have ( -\frac{2}{9} ) divided by ( -1 ). It's going to be positive ( \frac{2}{9} ).

So our sign right over here is going to be positive. And this is going to be the case in general when we're dealing with any negative value. Because if you take any negative value to the fifth power, it is going to be negative. And then you take that, the cube root of that, you're going to have negative, but then you have negative value divided by that, you're going to get a positive value.

So you can feel good that this test value is indicative of actually this entire interval. And if you're dealing with a positive value, well, that to the fifth power is going to be positive, and the cube root of that is still going to be positive. But then you're going to have ( -\frac{2}{9} ) divided by that positive value, so this is going to be negative.

So it is indeed the case that our concavity of ( g ) switches as we cross ( x = 0 ). We're concave upwards when ( x ) is less than ( 0 ); our second derivative is positive, and we are concave downwards when ( x ) is greater than ( 0 ).

Let me write that a little bit downwards: downwards when ( x ) is greater than ( 0 ). So we are switching concavity as we cross ( x = 0 ). And so this tells us that ( x ) ... so let's see, we are switching signs, switching. Let me say ( g'' ) is switching signs as we cross ( x = 0 ).

Our function is defined at ( x = 0 ) and it is defined at ( x = 0 ). So we have an inflection point at ( x = 0 ).

If you're familiar with the graph of the cube root, you would indeed see an inflection point at that point. So there we go: he was wrong in step three; there actually is an inflection point. It's not when the second derivative is equal to zero; it's actually where the second derivative is undefined.