### Video Transcript

Does the mean value theorem apply for the function π¦ is equal to two times the tan of three ππ₯ 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. Letβs 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 itβs differentiable on the open interval from π to π. In our case, our function π of π₯ is two times the tan of three ππ₯. 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.

Letβs 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, letβs start by considering the domain of the function the tan of π. And, of course, the tan of π wonβt 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 weβre 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. Itβs 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. Itβs not defined when π₯ is equal to three over two. So it canβt 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 ππ₯ over the closed interval from one to three.