# Question Video: Checking If We Can Apply the Mean Value Theorem to a Trigonometric Function Mathematics • Higher Education

Does the mean value theorem apply for the function π¦ = 2 tan 3ππ₯ over the interval [1, 3]?

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### Video Transcript

Does the mean value theorem apply for the function π¦ is equal to two times the tan of three π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. Letβs start by recalling what the mean value theorem tells us. The mean value theorem says if π is a function which is continuous on the closed interval from π to π and π is differentiable on the open interval from π to π. Then there exists some value of π in the open interval from π to π such that π prime of π is equal to π evaluated at π minus π evaluated at π divided by π minus π.

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 π is continuous on the closed interval from π to π. And we need to show that itβs differentiable on the open interval from π to π. In our case, our function π of π₯ is two times the tan of three ππ₯. Our value of π is one and our value of π is three. So we have two things to check. We need to check that our function π of π₯ is continuous on the closed interval from one to three. And we need to check that our function π of π₯ is differentiable on the open interval from one to three.

Letβs start by checking that two tan of three π 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 π of π₯ is defined for all values of π₯ between one and three. To do this, letβs start by considering the domain of the function the tan of π. And, of course, the tan of π wonβt be defined whenever we have a vertical asymptote or alternatively whenever the cos of π is equal to zero. And we know this happens when π is equal to π over two plus π times π where π is any integer.

But then this leaves us with a problem. We can see that weβre taking the tan of three π 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 π of π₯, we get two times the tan of three π multiplied by three over two. And this is where our problem lies. Three π multiplied by three over two can be simplified to give us nine π by two. And the tan of nine π by two is not defined. We have a vertical asymptote at this point.

One way of seeing this is nine π by two is equal to π by two plus four π. Itβs when π is equal to four. So if we wrote the tan of nine π by two as the sin of nine π by two divided by the cos of nine π by two, then, remember, we can add and subtract multiples of two π from our inputs. So the cos of nine π by two is just the cos of π by two. And we know this is equal to zero. So our function π 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 π of π₯ fails the first prerequisite for the mean value theorem. Itβs not defined when π₯ is equal to three over two. So it canβt possibly be continuous at this point. Therefore, we were able to show the mean value theorem does not apply to the function π¦ is equal to two times the tan of three ππ₯ over the closed interval from one to three.