Question Video: Determining Whether the Mean Value Theorem Can Be Applied for Trignometric Functions Mathematics • Higher Education

Does the mean value theorem apply for the function 𝑦 = 3π‘₯ sin 2πœ‹π‘₯ over the interval [0, 2] ?

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Video Transcript

Does the mean value theorem apply for the function 𝑦 is equal to three π‘₯ times the sin of two πœ‹π‘₯ 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. Let’s 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 𝑓 which is continuous on a closed interval from π‘Ž to 𝑏 and this function 𝑓 is differentiable on the open interval from π‘Ž to 𝑏. Then there must exist a 𝑐 on the open interval from π‘Ž to 𝑏 such that 𝑓 prime evaluated at 𝑐 is equal to 𝑓 evaluated at 𝑏 minus 𝑓 evaluated at π‘Ž divided by 𝑏 minus π‘Ž.

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

So we’ll set our function 𝑓 of π‘₯ to be three π‘₯ times the sin of two πœ‹π‘₯, π‘Ž to be equal to zero, and our value of 𝑏 equal to two. So we need to check our two conditions. We need to check 𝑓 of π‘₯ is continuous on the closed interval from zero to two. And we need to check our function 𝑓 of π‘₯ is differentiable on the open interval from zero to two.

So let’s start with the first condition. We’ll start by noticing our function 𝑓 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 πœ‹π‘₯ is defined for all real values of π‘₯. So three π‘₯ is continuous, and the sin of two πœ‹π‘₯ is also continuous.

So our function 𝑓 of π‘₯ is the product of two continuous functions. And we know the product of two continuous functions is itself continuous. This means we’ve shown that our function three π‘₯ times the sin of two πœ‹π‘₯ 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 πœ‹π‘₯ differentiable on the open interval from zero to two? There’s 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 πœ‹π‘₯ 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, we’ll also show how to do this by finding an expression for 𝑓 prime of π‘₯. Since our function 𝑓 of π‘₯ is the product of two functions, we’ll do this by using the product rule. We recall the product rule tells us the derivative of 𝑒 times 𝑣 is equal to 𝑒 prime 𝑣 plus 𝑣 prime 𝑒. We’ll set 𝑒 to be our first factor of three π‘₯ and 𝑣 to be our second factor the sin of two πœ‹π‘₯.

We now need to find expressions for 𝑒 prime and 𝑣 prime. We’ll start with 𝑒 prime. That’s 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 𝑣 with respect to π‘₯, we recall for any constant π‘˜, the derivative of the sin of π‘˜π‘₯ with respect to π‘₯ is equal to π‘˜ times the cos of π‘˜π‘₯. In our case, our value of π‘˜ is equal to two πœ‹. So using this, we get 𝑣 prime is equal to two πœ‹ times the cos of two πœ‹π‘₯.

Substituting in our values of 𝑒, 𝑣, 𝑒 prime, and 𝑣 prime into the product rule, we get that 𝑓 prime of π‘₯ is equal to three times the sin of two πœ‹π‘₯ plus two πœ‹ cos of two πœ‹π‘₯ 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 it’s clear this is true for our expression 𝑓 prime of π‘₯. For example, the sin of two πœ‹π‘₯ exists for all real values of π‘₯. This means we can multiply it by three.

Similarly, the cos of two πœ‹π‘₯ exists for all real values of π‘₯. We can multiply this by two πœ‹ 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 we’ve shown both of our conditions for the mean value theorem are true. And so we’ve shown that you can use the mean value theorem on the function 𝑦 is equal to three π‘₯ times the sin of two πœ‹π‘₯ over the closed interval from zero to two.

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