# Video: APCALC02AB-P1A-Q06-860181024643

Find the value of π that satisfies Rolleβs theorem for π(π₯) = 3π₯Β³ β 12π₯ on the interval [0, 2].

02:25

### Video Transcript

Find the value of π that satisfies Rolleβs theorem for π of π₯ equals three π₯ cubed minus 12π₯ on the closed interval zero to two.

Remember Rolleβs theorem says that if we have a function π of π₯, which is continuous on the closed interval π to π and differential on the open interval π to π, and π of π equals π of π. Then there exists at least one point π in the open interval π to π, for which π prime of π, the derivative of the function evaluated at π, is equal to zero. By comparing our question to the definition for Rolleβs theorem, we see that we can let π be equal to zero and π be equal to two. We know that for Rolleβs theorem to be true, π of π must be equal to π of π. So itβs sensible to check that this is the case. Letβs evaluate π of zero and π of two.

π of zero is three times zero cubed minus 12 times zero, which is simply zero. And π of two is three times two cubed minus 12 times two, which is also zero. So this part of the theorem is indeed correct. We can also say that our function π of π₯ is continuous and differentiable, since all polynomial functions are continuous and differentiable on the entire set of real numbers. And so weβre looking to find the value of π such that π prime of π is equal to zero.

Next then, weβll find the derivative of our function π prime of π₯. The derivative of three π₯ cubed is three times three π₯ squared. And the derivative of negative 12π₯ is negative 12. So we can say that π prime of π₯ is equal to nine π₯ squared minus 12. Replacing π₯ with π and we see that π prime of π is nine π squared minus 12. Weβre interested in the point at which this function is equal to zero. So letβs set nine π squared minus 12 equal to zero and solve for π.

We add 12 to both sides of our equation and then divide through by nine. And, of course, 12 over nine simplifies to four-thirds. Our next step is to take the square root of both sides of this equation. Usually, weβd look to take both a positive and negative square root of four-thirds. But since weβre interested in values of π₯ between zero and two, weβre just interested in the positive square root this time.

And we found our solution. The value of π that satisfies Rolleβs theorem in this case is the square root of four-thirds.