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Unformatted text preview: MATH 1A FALL 2009 THE MEAN VALUE THEOREM THE MEAN VALUE THEOREM HYP: ඃ IS CONTINUOUS ON THE CLOSED INTERVAL [ඃ, ඃ AND ඃ IS DIFFERENTIABLE ON THE OPEN INTERVAL (ඃ, ඃ ) CLAIM: THERE EXISTS A NUMBER ඃ IN (ඃ, ඃ) SUCH THAT ඃ ′ (ඃ) = ඃ (ඃ )−ඃ (ඃ ) ඃ −ඃ Sketch: We know we are going to use Rolle’s Theorem, so let’s try to rewrite the conclusion in terms of the conclusion of Rolle’s Theorem: Claim: There exists a number ඃ in (ඃ, ඃ) such that ඃ′ (ඃ) = ඃ ′ (ඃ) − ඃ(ඃ) − ඃ(ඃ) = 0. ඃ − ඃ Now this means we need a new function, say ඃ, and we will make ඃ′ (ඃ) = 0. If ඃ′ (ඃ) = ඃ ′ (ඃ) −
ඃ (ඃ )−ඃ (ඃ ) , then we try ඃ(ඃ ) ඃ −ඃ = ඃ(ඃ ) − ඃ (ඃ )−ඃ (ඃ ) ඃ because this will satisfy ඃ′ (ඃ ) ඃ −ඃ = ඃ ′ (ඃ) − ඃ (ඃ )−ඃ (ඃ ) . ඃ −ඃ Before we can apply Rolle’s Theorem to our function ඃ, we must check that ඃ(ඃ) = ඃ(ඃ). Let’s see: ඃ(ඃ ) = ඃ(ඃ ) −
ඃ (ඃ )−ඃ (ඃ ) ඃ, so ඃ(ඃ) would be ඃ(ඃ) ඃ −ඃ = ඃ(ඃ) − ඃ (ඃ )−ඃ (ඃ ) ඃ. and ඃ(ඃ) ඃ −ඃ = ඃ(ඃ) − ඃ (ඃ )−ඃ (ඃ ) ඃ Let’s check if this means ඃ(ඃ) ඃ −ඃ = ඃ(ඃ) : ඃ(ඃ) − ඃ(ඃ) ඃ(ඃ) − ඃ (ඃ) ඃ = ඃ(ඃ) − ඃ ඃ − ඃ ඃ − ඃ (ඃ − ඃ)ඃ(ඃ) − (ඃ(ඃ) − ඃ(ඃ))ඃ (ඃ − ඃ)ඃ(ඃ) − (ඃ(ඃ) − ඃ(ඃ))ඃ = ඃ − ඃ ඃ − ඃ ඃ (ඃ) − −ඃඃ (ඃ) + ඃඃ(ඃ) = ඃඃ(ඃ) − ඃඃ (ඃ) Good! So it turns out that if we use ඃ(ඃ ) = ඃ (ඃ ) −
ඃ (ඃ )−ඃ (ඃ ) ඃ , we will have ඃ(ඃ) ඃ −ඃ = ඃ(ඃ), and when we apply Rolle’s Theorem, there exists a ඃ in (ඃ, ඃ) such that ඃ′ (ඃ) = 0 and that is the same as ඃ ′ (ඃ) −
ඃ (ඃ )−ඃ (ඃ ) ඃ −ඃ = 0. And of course, since ඃ is continuous on the closed interval [ඃ, ඃ and ඃ is differentiable on the open interval (ඃ, ඃ), ඃ is continuous on the closed interval [ඃ, ඃ and ඃ is differentiable on the open interval (ඃ, ඃ). Okay: now the Write‐up… Proof: Create the new function ඃ(ඃ ) = ඃ (ඃ ) − ඃ (ඃ )−ඃ (ඃ ) ඃ −ඃ ඃ . Then ඃ satisfies the hypotheses of Rolle’s Theorem because 1. ඃ is the sum of ඃ and a polynomial, so since ඃ is continuous on the closed interval [ඃ, ඃ by hypothesis, so is ඃ. 2. ඃ is the sum of ඃ and a polynomial, so since ඃ is differentiable on the open interval (ඃ, ඃ) by hypothesis, so is ඃ. 3. ඃ(ඃ) = ඃ(ඃ) since ඃ(ඃ) = ඃ (ඃ) − = ඃ (ඃ) − ඃ(ඃ) ඃ ඃ − ඃ (ඃ − ඃ)ඃ(ඃ) − +ඃ(ඃ) − ඃ (ඃ),ඃ ඃ − ඃ ඃඃ (ඃ) − ඃඃ(ඃ) − ඃඃ (ඃ) + ඃඃ (ඃ) = ඃ − ඃ −ඃඃ (ඃ) + ඃඃ (ඃ) = ඃ − ඃ ඃඃ (ඃ) − ඃඃ (ඃ) − ඃඃ (ඃ) + ඃඃ (ඃ) = ඃ − ඃ (ඃ − ඃ)ඃ(ඃ) − (ඃ(ඃ) − ඃ(ඃ))ඃ = ඃ − ඃ ඃ (ඃ) − ඃ (ඃ) = ඃ (ඃ) − ඃ ඃ − ඃ = ඃ(ඃ). So, there exists a number ඃ in (ඃ, ඃ) such that ඃ′ (ඃ) = 0. Now, ඃ′ (ඃ ) = ඃ ′ (ඃ ) −
ඃ (ඃ )−ඃ (ඃ ) ඃ −ඃ ඃ (ඃ )−ඃ (ඃ ) ඃ −ඃ , so ඃ′ (ඃ) = ඃ ′ (ඃ ) − . So this means there exists a number ඃ in (ඃ, ඃ) such that ඃ ′ (ඃ) − ඃ (ඃ )−ඃ (ඃ ) ඃ −ඃ = 0. And finally, this means there exists a ඃ (ඃ )−ඃ (ඃ ) ඃ −ඃ number ඃ in (ඃ, ඃ) such that ඃ ′ (ඃ) = . ...
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This note was uploaded on 06/26/2010 for the course MATH 1A taught by Professor Wilkening during the Fall '08 term at University of California, Berkeley.
 Fall '08
 WILKENING

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