### Video Transcript

At how many points on the closed interval from π by two to three π by two does π of π₯ equals two cos of three π₯ minus cos of two π₯ satisfy the mean value theorem?

Letβs begin by simply quoting the mean value theorem itself. Let π be a function that satisfies the following criteria. (1) π is continuous on the closed interval from π to π. And (2) π is differentiable on the open interval from π to π. If π satisfies these criteria, then there exists a number π in the closed interval π to π such that π prime of π, the derivative of the function evaluated at π, is equal to π of π minus π of π over π minus π. Our function π of π₯ is given by two cos of three π₯ minus cos of two π₯. Cos of π₯ is both continuous and differentiable over its entire domain. And so, this in turn means that two cos of three π₯ minus cos of two π₯ must be continuous and differentiable over its entire domain, which happens to be here between π by two and three π by two radians.

And so, we turn to the third point. There exists some number π in the open interval π to π such that π prime of π is equal to π of π minus π of π over π minus π. Well, we can work out π of π minus π of π over π minus π quite easily. We know π is π by two and π is three π by two. So, we evaluate π of three π by two minus π of π by two over three π by two minus π by two. π of three π by two is found by substituting π₯ equals three π by two into our original function. And itβs equal to one. Similarly, π of π by two is found by substituting π₯ is equal to π by two into our original function, and we also get one. So, we get one minus one over π, which is simply zero.

And so, we want to find the number of points on the closed interval from π by two to three π by two where the derivative of our function is equal to zero. So, letβs differentiate two cos of three π₯ minus cos of two π₯ with respect to π₯. We do this term by term. Now, we can quote the general result for the derivative of cos of ππ₯. Itβs negative π sin of ππ₯ for real constants π. And so, the derivative of two cos of three π₯ is two times negative three sin of three π₯, which is negative six sin of three π₯. And then when we differentiate cos of two π₯, we get negative two sin of two π₯. And so, when we simplify, we find π prime of π₯ is equal to negative six sin of three π₯ plus two sin of two π₯.

Now, of course, weβre looking to find π such that π prime of π is equal to π of π minus π of π over π minus π. So, π prime of π is negative six sin of three π plus two sin of two π. And so, weβre looking for the values of π such that this expression is equal to zero. Now, this is not a particularly nice equation to solve by hand, but we can use a graphical calculator to help us. A sketch of the curve might look a little something like this. Negative six sin three π plus two sin two π is equal to zero at all the points where the curve crosses the π₯-axis. The first few of these are at zero, 0.939, 2.008, π, 4.275, and 5.344.

But of course, weβre looking for points on the closed interval from π by two to three π by two radians. Thatβs 1.57 to 4.712. There are three points that lie in this interval. Those are 2.008, π, and 4.275. And so, the answer is three. There are three points on the closed interval π by two to three π by two where the function two cos of three π₯ minus cos of two π₯ satisfies the mean value theorem.