Video: Checking If We Can Apply the Mean Value Theorem to a Trigonometric Function

Does the mean value theorem apply for the function 𝑦 = 2 tan 3πœ‹π‘₯ over the interval [1, 3]?

03:19

Video Transcript

Does the mean value theorem apply for the function 𝑦 is equal to two times the tan of three πœ‹π‘₯ over the closed interval from one to three?

The question gives us a trigonometric function. And it wants us to determine whether we can use the mean value theorem on this function over the closed interval from one to three. Let’s start by recalling what the mean value theorem tells us. The mean value theorem says if 𝑓 is a function which is continuous on the closed interval from π‘Ž to 𝑏 and 𝑓 is differentiable on the open interval from π‘Ž to 𝑏. Then there exists some value of 𝑐 in the open interval from π‘Ž to 𝑏 such that 𝑓 prime of 𝑐 is equal to 𝑓 evaluated at 𝑏 minus 𝑓 evaluated at π‘Ž divided by 𝑏 minus π‘Ž.

So to use the mean value theorem on a function over an interval, there are two prerequisites. First, we need to show that our function 𝑓 is continuous on the closed interval from π‘Ž to 𝑏. And we need to show that it’s differentiable on the open interval from π‘Ž to 𝑏. In our case, our function 𝑓 of π‘₯ is two times the tan of three πœ‹π‘₯. Our value of π‘Ž is one and our value of 𝑏 is three. So we have two things to check. We need to check that our function 𝑓 of π‘₯ is continuous on the closed interval from one to three. And we need to check that our function 𝑓 of π‘₯ is differentiable on the open interval from one to three.

Let’s start by checking that two tan of three πœ‹ by π‘₯ is continuous on the closed interval from one to three. First, we see that our function is a trigonometric function. And we know that trigonometric functions are continuous across their entire domain. So to check our first prerequisite, we just need to check that 𝑓 of π‘₯ is defined for all values of π‘₯ between one and three. To do this, let’s start by considering the domain of the function the tan of πœƒ. And, of course, the tan of πœƒ won’t be defined whenever we have a vertical asymptote or alternatively whenever the cos of πœƒ is equal to zero. And we know this happens when πœƒ is equal to πœ‹ over two plus 𝑛 times πœ‹ where 𝑛 is any integer.

But then this leaves us with a problem. We can see that we’re taking the tan of three πœ‹ times π‘₯. So we can consider what would happen if π‘₯ was equal to three over two. Substituting π‘₯ is equal to three over two into our function 𝑓 of π‘₯, we get two times the tan of three πœ‹ multiplied by three over two. And this is where our problem lies. Three πœ‹ multiplied by three over two can be simplified to give us nine πœ‹ by two. And the tan of nine πœ‹ by two is not defined. We have a vertical asymptote at this point.

One way of seeing this is nine πœ‹ by two is equal to πœ‹ by two plus four πœ‹. It’s when 𝑛 is equal to four. So if we wrote the tan of nine πœ‹ by two as the sin of nine πœ‹ by two divided by the cos of nine πœ‹ by two, then, remember, we can add and subtract multiples of two πœ‹ from our inputs. So the cos of nine πœ‹ by two is just the cos of πœ‹ by two. And we know this is equal to zero. So our function 𝑓 of π‘₯ is not defined when π‘₯ is equal to three over two. However, three over two is in our closed interval from one to three.

Therefore, our function 𝑓 of π‘₯ fails the first prerequisite for the mean value theorem. It’s not defined when π‘₯ is equal to three over two. So it can’t possibly be continuous at this point. Therefore, we were able to show the mean value theorem does not apply to the function 𝑦 is equal to two times the tan of three πœ‹π‘₯ over the closed interval from one to three.

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