Video: Applications of the Mean Value Theorem

Which is the only function that satisfies the mean value theorem on the interval specified? [A] 𝑓(π‘₯) = (π‘₯Β² βˆ’ 2)/(π‘₯ βˆ’ 1) on [1, 3] [B] 𝑓(π‘₯) = π‘₯Β² βˆ’ (π‘₯/2) on [βˆ’2, 1] [C] 𝑓(π‘₯) = π‘₯ βˆ’ (1/π‘₯) on [βˆ’2, 2] [D] 𝑓(π‘₯) = 2/(π‘₯Β² βˆ’ 9) on [βˆ’1, 3]

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Video Transcript

Which is the only function that satisfies the mean value theorem on the interval specified? Is it a) 𝑓 of π‘₯ equals π‘₯ squared minus two over π‘₯ minus one on the closed interval one to three? b) 𝑓 of π‘₯ equals π‘₯ squared minus π‘₯ over two on the closed interval negative two to one. Is it c) 𝑓 of π‘₯ equals π‘₯ minus one over π‘₯ on the closed interval negative two to two? Or d) 𝑓 of π‘₯ equals two over π‘₯ squared minus nine on the closed interval negative one to three.

Let’s begin simply by quoting the mean value theorem. The mean value theorem says that if 𝑓 satisfies the two criteria, it’s continuous on the closed interval π‘Ž to 𝑏 and differentiable on the open interval π‘Ž to 𝑏. Then there exists a number 𝑐 in the interval π‘Ž to 𝑏 such that 𝑓 prime of 𝑐 is equal to 𝑓 of 𝑏 minus 𝑓 of π‘Ž over 𝑏 minus π‘Ž. So we’ll begin simply by checking for continuity and differentiability of each of our functions on their intervals.

Let’s begin with the function 𝑓 of π‘₯ equals π‘₯ squared minus two over π‘₯ minus one. A big problem with continuity is when we’re dealing with fractional functions. If the denominator of the fraction is equal to zero at any point on the interval, that means that the function is not defined here. And so it absolutely cannot be continuous. So for what value of π‘₯ is π‘₯ minus one equal to zero? Well, solving for π‘₯ and we find that gives us π‘₯ is equal to one. Now, one is in the closed interval from one to three. So that tells us that our function 𝑓 of π‘₯ equals π‘₯ squared minus two over π‘₯ minus one is not continuous on the closed interval one to three. And so we can instantly disregard a.

We’ll now consider b. b does have a fractional term in it. But the denominator of that term is two, so there’s no way that cannot be equal to zero. In fact, the function itself is a polynomial function. And we know that polynomials are continuous and differentiable on their entire domain. And so b does satisfy the first two criteria. 𝑓 is continuous on the closed interval π‘Ž to 𝑏 and differentiable on the open interval π‘Ž to 𝑏.

We’ll check the third part in a moment. Let’s check c and d next. In c, our function does have a fractional term. And we can actually see that if π‘₯ is equal to zero, the function is not continuous at this point. Well, zero is in the closed interval negative two to two. And so 𝑐 is not continuous on that closed interval. And what about d? Well, once again, we need to establish the values of π‘₯ such that π‘₯ squared minus nine will be equal to zero.

Well, by adding nine and taking the positive and negative square root, we find that π‘₯ is equal to positive or negative three. Three is in the closed interval negative one to three, so we can see that the function would not be continuous on that closed interval. And so it’s safe to assume that the answer is b. But we are going to check the second part of the mean value theorem. This says that 𝑓 prime of 𝑐 is equal to 𝑓 of 𝑏 minus 𝑓 of π‘Ž over 𝑏 minus π‘Ž.

Let’s begin by working out 𝑓 prime of π‘₯. We’re going to differentiate our function with respect to π‘₯. We can do this term by term. The derivative of π‘₯ squared is two times π‘₯; it’s two π‘₯. And the derivative of negative π‘₯ over two is negative one-half. 𝑓 prime of 𝑐 is therefore two 𝑐 minus one-half. If we let π‘Ž be equal to negative two and 𝑏 be equal to one, we can see that 𝑓 of 𝑏 minus 𝑓 of π‘Ž over 𝑏 minus π‘Ž is 𝑓 of one minus 𝑓 of negative two over one minus negative two. That’s one squared minus a half minus negative two squared minus negative two over two all over three, which is negative three over two.

According to the mean value theorem, there is some 𝑐 in our open interval from negative two to one such that two 𝑐 minus a half is equal to negative three over two. Let’s solve for 𝑐 by adding a half to both sides. Well, negative three over two plus a half is negative one. So two 𝑐 equals negative one. And then, we divide through by two. And then we divide through to get 𝑐 is equal to negative one-half. Well, that is indeed in the open interval from negative two to one.

And so the answer is b. The only function that satisfies the mean value theorem on the interval specified is 𝑓 of π‘₯ equals π‘₯ squared minus π‘₯ over two.

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