Proving with mean value theorem

AI Thread Summary
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.
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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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Well, what does the mean value theorem say, and what does it imply when you use it as hinted?
 
I'd also keep in mind the definition of a derivative.

--J
 
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