Video: APCALC05AB-P1B-Q38-519192503705

At how many points on the interval [πœ‹/2, 3πœ‹/2] does 𝑓(π‘₯) = 2 cos 3π‘₯ βˆ’ cos 2π‘₯ satisfy the mean value theorem?

03:15

Video Transcript

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

Let’s begin by simply quoting the mean value theorem itself. Let 𝑓 be a function that satisfies the following criteria. (1) 𝑓 is continuous on the closed interval from π‘Ž to 𝑏. And (2) 𝑓 is differentiable on the open interval from π‘Ž to 𝑏. If 𝑓 satisfies these criteria, then there exists a number 𝑐 in the closed interval π‘Ž to 𝑏 such that 𝑓 prime of 𝑐, the derivative of the function evaluated at 𝑐, is equal to 𝑓 of 𝑏 minus 𝑓 of π‘Ž over 𝑏 minus π‘Ž. Our function 𝑓 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œ‹ by two and three πœ‹ by two radians.

And so, we turn to the third point. There exists some number 𝑐 in the open 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. Well, we can work out 𝑓 of 𝑏 minus 𝑓 of π‘Ž over 𝑏 minus π‘Ž quite easily. We know π‘Ž is πœ‹ by two and 𝑏 is three πœ‹ by two. So, we evaluate 𝑓 of three πœ‹ by two minus 𝑓 of πœ‹ by two over three πœ‹ by two minus πœ‹ by two. 𝑓 of three πœ‹ by two is found by substituting π‘₯ equals three πœ‹ by two into our original function. And it’s equal to one. Similarly, 𝑓 of πœ‹ by two is found by substituting π‘₯ is equal to πœ‹ by two into our original function, and we also get one. So, we get one minus one over πœ‹, which is simply zero.

And so, we want to find the number of points on the closed interval from πœ‹ by two to three πœ‹ by two where the derivative of our function is equal to zero. So, let’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‘Žπ‘₯. It’s negative π‘Ž sin of π‘Žπ‘₯ for real constants π‘Ž. 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 𝑓 prime of π‘₯ is equal to negative six sin of three π‘₯ plus two sin of two π‘₯.

Now, of course, we’re looking to find 𝑐 such that 𝑓 prime of 𝑐 is equal to 𝑓 of 𝑏 minus 𝑓 of π‘Ž over 𝑏 minus π‘Ž. So, 𝑓 prime of 𝑐 is negative six sin of three 𝑐 plus two sin of two 𝑐. And so, we’re looking for the values of 𝑐 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 𝑐 plus two sin two 𝑐 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œ‹, 4.275, and 5.344.

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

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