# Video: The Mean Value Theorem

Mason is not convinced that the mean value theorem is true because, he says, the function π(π₯) = |π₯| has the property that if we take π = β2 and π = 2, we have (π(π) β π(π))/(π β π) = 0, and yet there is no point π₯ where πβ²(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 π of π₯ equals the absolute value of π₯ has the property that if we take π equals negative two and π equals two, we have π of π minus π of π over π minus π equals zero, and yet there is no point π₯ where π prime of π₯ equals zero. What is his error?

To use the mean value theorem, two things must be true. π of π₯ must be continuous over π, π. When we consider Masonβs function π of π₯ equals the absolute value of π₯, this statement is true. π of π₯ is continuous over π, π. It must also be true that our π of π₯ is differentiable over π, π.

And for Mason, π 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 π of π₯ equals the absolute value of π₯ is not differentiable over the interval π, π. And that means the mean value theorem cannot apply here.