Question Video: The Mean Value Theorem Mathematics • Higher Education

For the function π(π₯) = (π₯ β 1)βΈ, find all the possible values of π that satisfy the mean value theorem over the closed interval [0, 2].

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Video Transcript

For the function π of π₯ equals π₯ minus one to the eighth power, find all the possible values of π that satisfy the mean value theorem over the closed interval zero to two.

Remember, the mean value theorem says that if π is a function which is continuous over the closed interval π to π and differentiable at every point of the open interval π to π. Then thereβs a point π in that open interval, such that π prime of π is equal to π of π minus π of π over π minus π. Our π of π₯ is π₯ minus one to the eighth power. And our closed interval runs from zero to two. Weβre looking to find the value of π such that the derivative of our function evaluated at π is equal to π of π minus π of π over π minus π. Thatβs π of two minus π of zero over two minus zero. So weβll do two things. Weβll evaluate this quotient. And weβll also find an expression for the derivative of our function and evaluate it at π. π of two minus π of zero is two minus one to the eighth power minus zero minus one to the eighth power. And we obtain that π of two minus π of zero over two minus zero is zero divided by two, which is just zero.

Our next job is to find the derivative of our function. And weβll use the general power rule. This says that if π of π₯ is a differentiable function and π is π constant real number, such that π of π₯ is equal to π of π₯ to the power of π. Then the derivative of π of π₯, π prime of π₯, is equal to π times π of π₯ to the power of π minus one times the derivative of π of π₯. Thatβs π prime of π₯. The derivative of our function π₯ minus one to the eighth power is therefore eight times π₯ minus one to the seventh power multiplied by the derivative of π₯ minus one, which is one. And so we see that π prime of π₯ is eight times π₯ minus one to the seventh power.

We see that π prime of π is eight times π minus one to the seventh power. And remember, we found that π of π minus π of π over π minus π is zero. So weβre going to set this equal to zero and then solve for π. We divide by eight. And we see that π minus one to the seventh power is zero. And we can take the seventh root of both sides to see that π minus one is equal to zero, which means that π equals one is the only value of π that satisfies the mean value theorem. Notice it falls in the closed interval zero to two, as required.