# Video: Using the Mean Value Theorem to Verify Statements about the Derivative of a Function within an Interval

If the function π(π₯) = π₯Β² cos (πβπ₯), which of the following statements is a conclusion of the mean value theorem? [A] There exists π in (1, 9) such that πβ²(π) = β1/10. [B] There exists π in (0, 4) such that πβ²(π) = β4. [C] There exists π (1, 9) such that πβ²(π) = β10. [D] There exists π in (0, 4) such that πβ²(π) = 0.

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### Video Transcript

If the function π of π₯ is equal to π₯ squared multiplied by the cos of π square root of π₯, which of the following statements is a conclusion of the mean value theorem? Option a) there exists π in the open interval from one to nine such that the derivative of π evaluated at π is equal to negative one-tenth. Option b) there exists π in the open interval from zero to four such that the derivative of π evaluated at π is equal to negative four. Option c) there exists π in the open interval from one to nine such that the derivative of π evaluated at π is equal to negative 10. Or option d) there exists π in the open interval from zero to four such that the derivative of π evaluated at π is equal to zero.

The question wants us to apply the mean value theorem on the function π of π₯ which is equal to π₯ squared multiplied by the cos of π square root of π₯. We recall that the mean value theorem tells us that if we have a function π 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 π in the open interval from π to π such that the derivative of π evaluated at π is equal to π evaluated at π minus π evaluated at π all divided by π minus π.

We can see that the concluding statement in the mean value theorem gives us that there exists π in some open interval from π to π such that our derivative function π prime evaluated at π is equal to π of π minus π of π all divided by π minus π. If we look at the answers given to us in this question, weβre given two different open intervals. In options a and option c, thereβs the open interval from one to nine. And in options b and option d, thereβs the open interval from zero to four.

So, if we set π equal to one and π equal to nine in our mean value theorem, show that the prerequisites are true. Weβll get some expression for π prime evaluated at π, where π is in the open interval from one to nine. Similarly, if we set π equal to zero and π equal to four in our mean value theorem, show that all the prerequisites are true. Then, we would get that there exists π in the open interval from zero to four such that π prime of π is equal to π of four minus π of zero divided by four minus zero.

So, the next step is to check our prerequisites for the mean value theorem. First, we need that π is continuous on the closed interval from π to π. Second, we need that π is differentiable on the open interval from π to π. To check the continuity of our function π of π₯, we recall the following two properties of continuous functions. First, the composition of continuous functions will give us a continuous function. Second, the product of continuous functions will give us a continuous function.

We see that π of π₯ is the product of a polynomial, π₯ squared, and a composition function, the cos of π square root of π₯. We know that all polynomials are continuous, so π₯ squared is continuous. So, if we can show that the cos of π square root of π₯ is continuous, then we can conclude that the product π₯ squared multiplied by the cos of π square root of π₯, which is equal to π of π₯, is continuous. So, weβve shown that π₯ squared is continuous. Now, all we need to do is show that cos of π square root of π₯ is continuous.

Well, we know that the square root of π₯ on its own is continuous. Multiplying it by π is still continuous as well. And just the function on its own, the cos of π₯, is a trig function, which is also continuous. So, the cos of π square root of π₯ is the composition of continuous functions, which means that it itself is continuous. Therefore, since π of π₯ is the product of continuous functions, weβve shown that π of π₯ must be continuous on its domain.

So, we need to check the domain of our function π of π₯ to check where it is continuous. We see that π₯ squared is defined for any real number. The cos of a number is defined for any real number. Multiplying the number by π is defined for any real number. However, the square root of π₯ is only defined when π₯ is greater than or equal to zero. So, our function π of π₯ has a domain π₯ greater than or equal to zero. And therefore, itβs continuous when π₯ is greater than or equal to zero.

So, what weβve shown is that our first prerequisite of π being continuous on the closed interval from π to π is true in both cases. If π is equal to zero and π is equal to four or if π is equal to one and π is equal to nine. Now, we need to show that our second prerequisite is true. That π is differentiable on the open interval from π to π. To help us check that our function π is differentiable, we recall the product rule and the chain rule.

The product rule tells us that the derivative of π’ multiplied by π£ is equal to π’ prime π£ plus π£ prime π’. And the chain rule tells us that the derivative of the composition π’ of π£ of π₯ is equal to π£ prime of π₯ multiplied by π’ prime evaluated at π£ of π₯. Applying both of these rules and simplifying gives us that our derivative function π prime of π₯ is equal to two π₯ multiplied by the cos of π square root of π₯ minus ππ₯ multiplied by the square root of π₯ divided by two multiplied by the sin of π square root of π₯.

The only parts of this function which are not defined for all values of π₯ are when we take the square root of π₯. These are defined when π₯ is greater than or equal to zero. So, weβve shown that our derivative function π prime of π₯ exists on the open interval from zero to four and the open interval from one to nine. So, in either case, both prerequisites of our mean value theorem are true. So, now, we need to evaluate the expression given to us in the conclusion of the mean value theorem and compare this to the answers given to us in the question.

So, we can draw the conclusion from the mean value theorem that there exists π in the open interval from zero to four such that π prime evaluated at π is equal to π of four minus π of zero divided by four minus zero. Similarly, we can also conclude from the mean value theorem that there exists π in the open interval from one to nine such that π prime evaluated at π is equal to π of nine minus π of one all divided by nine minus one.

Evaluating our first expression gives us four squared multiplied by the cos of π square root of four minus zero squared multiplied by the cos of π square root of zero all divided by four. Our second term is multiplied by zero squared. So, thatβs just equal to zero. And we can simplify π multiplied by the square root of four to be equal to two π. This gives us four squared multiplied by the cos of two π divided by four. We can simplify one of the factors of four in our numerator and our denominator. And we know that the cos of two π is just equal to one.

So, our first conclusion to the mean value theorem tells us that there exists π in the open interval from zero to four such that π prime evaluated at π is equal to four. Similarly, we can evaluate the expression in our second conclusion to get nine squared multiplied by the cos of π square root of nine minus one squared multiplied by the cos of π square root of one all divided by nine minus one.

We can simplify the denominator to be equal to eight. We have that one squared is just equal to one. And the square root of one is just equal to one. We know that the cos of π is equal to negative one. So, instead of subtracting negative one, we can add one. We have that π multiplied by the square root of nine is equal to three π. And nine squared can be simplified to give us 81. This gives us 81 multiplied by the cos of three π plus one all divided by eight.

We have that the cos of three π is equal to negative one. So, we can simplify our numerator to be negative 81 plus one, which, of course, is just equal to negative 80. Finally, we can simplify negative 80 divided by eight to just be equal to negative 10. This gives us the conclusion to the mean value theorem applied to π of π₯ that there exists π in the open interval from one to nine such that π prime evaluated at π is equal to negative 10, which was our option c.