Proving with mean value theorem

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Start date Jan 4, 2005
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Mean Mean value theorem Theorem Value

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The discussion revolves around applying the Mean Value Theorem (MVT) to prove that |g(a+h)| < Mh^2 under the given conditions. It begins by noting that since g(a) = g'(a) = 0, the MVT can be applied to g' on the interval [a, a+k], leading to the conclusion that g'(k) = g''(c) for some c in (a, a+k). Given that |g''(x)| < M, it follows that |g'(k)| < Mh. Subsequently, integrating this result over the interval [0, h] leads to the final inequality |g(a+h)| < Mh^2. This proof effectively demonstrates the relationship between the second derivative and the behavior of the function over the specified interval.

Jan 4, 2005

trap

Suppose that g(a) = g'(a)=0 and |g''(x)| < M for all x in [a, a+h] (for some positive constant M). Show that |g(a+h)| < Mh^2.
(Hint: Let k be any number such that 0<= k <= h and apply Mean Value Theorem to g' on [a,a+k].)

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Jan 4, 2005

matt grime

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Well, what does the mean value theorem say, and what does it imply when you use it as hinted?

Jan 4, 2005

Justin Lazear

I'd also keep in mind the definition of a derivative.

--J

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