# Video: APCALC05AB-P1B-Q38-519192503705

At how many points on the interval [π/2, 3π/2] does π(π₯) = 2 cos 3π₯ β cos 2π₯ satisfy the mean value theorem?

03:15

### 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?

Letβs 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 itβs 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, letβs 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 ππ₯. Itβs negative π sin of ππ₯ 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, weβre 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, weβre 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, weβre looking for points on the closed interval from π by two to three π by two radians. Thatβs 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.