Video: APCALC02AB-P1A-Q06-860181024643

Find the value of 𝑐 that satisfies Rolle’s theorem for 𝑓(π‘₯) = 3π‘₯Β³ βˆ’ 12π‘₯ on the interval [0, 2].


Video Transcript

Find the value of 𝑐 that satisfies Rolle’s theorem for 𝑓 of π‘₯ equals three π‘₯ cubed minus 12π‘₯ on the closed interval zero to two.

Remember Rolle’s theorem says that if we have a function 𝑓 of π‘₯, which is continuous on the closed interval π‘Ž to 𝑏 and differential on the open interval π‘Ž to 𝑏, and 𝑓 of π‘Ž equals 𝑓 of 𝑏. Then there exists at least one point 𝑐 in the open interval π‘Ž to 𝑏, for which 𝑓 prime of 𝑐, the derivative of the function evaluated at 𝑐, is equal to zero. By comparing our question to the definition for Rolle’s theorem, we see that we can let π‘Ž be equal to zero and 𝑏 be equal to two. We know that for Rolle’s theorem to be true, 𝑓 of π‘Ž must be equal to 𝑓 of 𝑏. So it’s sensible to check that this is the case. Let’s evaluate 𝑓 of zero and 𝑓 of two.

𝑓 of zero is three times zero cubed minus 12 times zero, which is simply zero. And 𝑓 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 𝑓 of π‘₯ is continuous and differentiable, since all polynomial functions are continuous and differentiable on the entire set of real numbers. And so we’re looking to find the value of 𝑐 such that 𝑓 prime of 𝑐 is equal to zero.

Next then, we’ll find the derivative of our function 𝑓 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 𝑓 prime of π‘₯ is equal to nine π‘₯ squared minus 12. Replacing π‘₯ with 𝑐 and we see that 𝑓 prime of 𝑐 is nine 𝑐 squared minus 12. We’re interested in the point at which this function is equal to zero. So let’s set nine 𝑐 squared minus 12 equal to zero and solve for 𝑐.

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, we’d look to take both a positive and negative square root of four-thirds. But since we’re interested in values of π‘₯ between zero and two, we’re just interested in the positive square root this time.

And we found our solution. The value of 𝑐 that satisfies Rolle’s theorem in this case is the square root of four-thirds.

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