Video: The Mean Value Theorem

Mason is not convinced that the mean value theorem is true because, he says, the function 𝑓(π‘₯) = |π‘₯| has the property that if we take π‘Ž = βˆ’2 and 𝑏 = 2, we have (𝑓(𝑏) βˆ’ 𝑓(π‘Ž))/(𝑏 βˆ’ π‘Ž) = 0, and yet there is no point π‘₯ where 𝑓′(π‘₯) = 0. What is his error?

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Video Transcript

Mason is not convinced that the mean value theorem is true because he says, the function 𝑓 of π‘₯ equals the absolute value of π‘₯ has the property that if we take π‘Ž equals negative two and 𝑏 equals two, we have 𝑓 of 𝑏 minus 𝑓 of π‘Ž over 𝑏 minus π‘Ž equals zero, and yet there is no point π‘₯ where 𝑓 prime of π‘₯ equals zero. What is his error?

To use the mean value theorem, two things must be true. 𝑓 of π‘₯ must be continuous over π‘Ž, 𝑏. When we consider Mason’s function 𝑓 of π‘₯ equals the absolute value of π‘₯, this statement is true. 𝑓 of π‘₯ is continuous over π‘Ž, 𝑏. It must also be true that our 𝑓 of π‘₯ is differentiable over π‘Ž, 𝑏.

And for Mason, 𝑓 prime of the absolute value of π‘₯ equals π‘₯ over the square root of π‘₯ squared and π‘₯ cannot be equal to zero. The function 𝑓 of π‘₯ equals the absolute value of π‘₯ is not differentiable over the interval π‘Ž, 𝑏. And that means the mean value theorem cannot apply here.

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