Video: Finding the Values That Satisfy the Mean Value Theorem for a Cubic Function in a Given Interval

For the function π(π₯) = 3π₯Β³, find all possible values of π that satisfy the mean value theorem over the interval [1, 3].

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Video Transcript

For the function π of π₯ is equal to three π₯ cubed, find all possible values of π that satisfy the mean value theorem over the closed interval from one to three.

The question gives us a function π of π₯ is equal to three π₯ cubed. It wants us to find all values of π which satisfy the mean value theorem over the closed interval from one to three. We recall the mean value theorem tells us that if the function π of π₯ is continuous on the closed interval from π to π and itβs differentiable on the open interval from π to π. Then there must exist a π in the open interval from π to π such that our derivative function π prime evaluated at π is equal to π evaluated at π minus π evaluated at π all divided by π minus π.

The question wants us to use the mean value theorem on the closed interval from one to three. So, weβll set π equal to one and π equal to three. Since the question wants us to find all of the values of π which satisfy the mean value theorem, we need π to be in our open interval from π to π. And we need π prime evaluated at π to be equal to π of π minus π of π over π minus π.

Since we want π to be in the open interval from π to π and we have π is one and π is three, we have that one is less than π is less than three. And weβre looking for all π in the open interval from one to three such that π prime of π is equal to π of three minus π of one over three minus one. We can evaluate π at three and π at one by substituting π₯ is equal to three and π₯ is equal to one into our formula for π of π₯. This gives us three multiplied by three cubed minus three multiplied by one cubed. And we can simplify our denominator to be two.

We can calculate this to give us 81 minus three divided by two, which is equal to 39. Now, letβs calculate the derivative function π prime of π₯. This is the derivative with respect to π₯ of three π₯ cubed. We know, for constants π and π, we can differentiate ππ₯ to the πth power with respect to π₯ by multiplying by our exponent and then reducing the exponent by one.

To apply this to three π₯ cubed, we multiply by our exponent of three and reduce the exponent by one. This gives us that π prime of π₯ is equal to nine π₯ squared. We remember weβre looking for all values of π such that one is less than π is less than three and π prime evaluated at π is equal to 39. By using that π prime of π₯ is equal to nine π₯ squared, weβre looking for values of π such that nine π squared is equal to 39.

We divide both sides of our equation by nine to get that π squared is equal to 39 divided by nine. Solving this normally by taking square roots gives us that π is equal to plus or minus the square root of 39 divided by three. However, we remember weβre only looking for values of π which are in the open interval from one to three. So, we only want the positive square root. Therefore, weβve shown for the function π of π₯ is equal to three π₯ cubed, all values of π which satisfy the mean value theorem on the closed interval from one to three is just the value π is equal to the square root of 39 divided by three.