Video: The Mean Value Theorem

For the function 𝑓(π‘₯) = (π‘₯ βˆ’ 1)⁸, find all the possible values of 𝑐 that satisfy the mean value theorem over the closed interval [0, 2].

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Video Transcript

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

Remember, the mean value theorem says that if 𝑓 is a function which is continuous over the closed interval π‘Ž to 𝑏 and differentiable at every point of the open interval π‘Ž to 𝑏. Then there’s a point 𝑐 in that open interval, such that 𝑓 prime of 𝑐 is equal to 𝑓 of 𝑏 minus 𝑓 of π‘Ž over 𝑏 minus π‘Ž. Our 𝑓 of π‘₯ is π‘₯ minus one to the eighth power. And our closed interval runs from zero to two. We’re looking to find the value of 𝑐 such that the derivative of our function evaluated at 𝑐 is equal to 𝑓 of 𝑏 minus 𝑓 of π‘Ž over 𝑏 minus π‘Ž. That’s 𝑓 of two minus 𝑓 of zero over two minus zero. So we’ll do two things. We’ll evaluate this quotient. And we’ll also find an expression for the derivative of our function and evaluate it at 𝑐. 𝑓 of two minus 𝑓 of zero is two minus one to the eighth power minus zero minus one to the eighth power. And we obtain that 𝑓 of two minus 𝑓 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 we’ll use the general power rule. This says that if 𝑔 of π‘₯ is a differentiable function and 𝑛 is π‘Ž constant real number, such that 𝑓 of π‘₯ is equal to 𝑔 of π‘₯ to the power of 𝑛. Then the derivative of 𝑓 of π‘₯, 𝑓 prime of π‘₯, is equal to 𝑛 times 𝑔 of π‘₯ to the power of 𝑛 minus one times the derivative of 𝑔 of π‘₯. That’s 𝑔 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 𝑓 prime of π‘₯ is eight times π‘₯ minus one to the seventh power.

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

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