### Video Transcript

Does the mean value theorem apply for the function π¦ is equal to three π₯ times the sin of two ππ₯ over the closed interval from zero to two?

The question gives us a function which is a product between a linear function and a trigonometric function. It wants us to determine whether we can apply the mean value theorem to this function on the closed interval from zero to two. Letβs start by recalling what the mean value theorem tells us about a function over an interval.

The mean value theorem says if we have a function π which is continuous on a closed interval from π to π and this function π is differentiable on the open interval from π to π. Then there must exist a π on the open interval from π to π such that π prime evaluated at π is equal to π evaluated at π minus π evaluated at π divided by π minus π.

We can see there are two conditions on our function π to use the mean value theorem. First, our function π must be continuous on the closed interval from π to π. Second, π must be differentiable on the open interval from π to π. We want to check if we can use the mean value theorem on π¦ is equal to three π₯ times the sin of two ππ₯ over the closed interval from zero to two.

So weβll set our function π of π₯ to be three π₯ times the sin of two ππ₯, π to be equal to zero, and our value of π equal to two. So we need to check our two conditions. We need to check π of π₯ is continuous on the closed interval from zero to two. And we need to check our function π of π₯ is differentiable on the open interval from zero to two.

So letβs start with the first condition. Weβll start by noticing our function π of π₯ is the product of a linear function and a trigonometric function. And we know a few things about these functions. We know all linear functions are continuous for all real numbers. We also know that trigonometric functions are continuous across their entire domain. And we know the sin of two ππ₯ is defined for all real values of π₯. So three π₯ is continuous, and the sin of two ππ₯ is also continuous.

So our function π of π₯ is the product of two continuous functions. And we know the product of two continuous functions is itself continuous. This means weβve shown that our function three π₯ times the sin of two ππ₯ is continuous for all real values of π₯. In particular, this means it must be continuous on the closed interval from zero to two.

We now need to check our second condition. Is three π₯ times the sin of two ππ₯ differentiable on the open interval from zero to two? Thereβs a few different ways of doing this. For example, we know three π₯ is differentiable for all real values of π₯, and the sin of two ππ₯ is also differentiable for all real values of π₯. This is actually enough to just say that their product will also be differentiable for all real values of π₯. And so it will be differentiable on the open interval from zero to two.

However, weβll also show how to do this by finding an expression for π prime of π₯. Since our function π of π₯ is the product of two functions, weβll do this by using the product rule. We recall the product rule tells us the derivative of π’ times π£ is equal to π’ prime π£ plus π£ prime π’. Weβll set π’ to be our first factor of three π₯ and π£ to be our second factor the sin of two ππ₯.

We now need to find expressions for π’ prime and π£ prime. Weβll start with π’ prime. Thatβs the derivative of the linear function three π₯. Since this is a linear function, the derivative will be the coefficient of π₯, which in this case is three. To differentiate our function π£ with respect to π₯, we recall for any constant π, the derivative of the sin of ππ₯ with respect to π₯ is equal to π times the cos of ππ₯. In our case, our value of π is equal to two π. So using this, we get π£ prime is equal to two π times the cos of two ππ₯.

Substituting in our values of π’, π£, π’ prime, and π£ prime into the product rule, we get that π prime of π₯ is equal to three times the sin of two ππ₯ plus two π cos of two ππ₯ multiplied by three π₯. And we could start rearranging and simplifying this expression. However, remember, we only need to show that our function is differentiable on the open interval from zero to two. And itβs clear this is true for our expression π prime of π₯. For example, the sin of two ππ₯ exists for all real values of π₯. This means we can multiply it by three.

Similarly, the cos of two ππ₯ exists for all real values of π₯. We can multiply this by two π and multiply this by three π₯. Finally, we can just add these together. This will exist for all real values of π₯. In particular, this means our function is differentiable on the open interval from zero to two. And this means weβve shown both of our conditions for the mean value theorem are true. And so weβve shown that you can use the mean value theorem on the function π¦ is equal to three π₯ times the sin of two ππ₯ over the closed interval from zero to two.