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Homework Statement
Let f: U -> R where U is an open subset of R. Suppose that for some c in U, f'(c) = 0 and f''(c) < 0. Proof that the restriction of f to some open ball of center c attains a maximum at c.The attempt at a solution
Let e > 0. Then for some d' > 0, if |x - c| < d' then |f(x) - f(c)| ≤ e|x - c| since f'(c) = 0. Let f''(c) = -b for some positive b. Then for some d'' > 0, if |x - c| < d'' then |f'(x) + b(x - c)| ≤ e|x - c|. Thus, for d = min{d', d''}, if |x - c| < d then
|f(x) - f(c)| ≤ e|x - c| and |f'(x) + b(x - c)| ≤ e|x - c|.
Of course, what I need is f(x) ≤ f(c). I don't see how to get that from the above. Any tips?
Let f: U -> R where U is an open subset of R. Suppose that for some c in U, f'(c) = 0 and f''(c) < 0. Proof that the restriction of f to some open ball of center c attains a maximum at c.The attempt at a solution
Let e > 0. Then for some d' > 0, if |x - c| < d' then |f(x) - f(c)| ≤ e|x - c| since f'(c) = 0. Let f''(c) = -b for some positive b. Then for some d'' > 0, if |x - c| < d'' then |f'(x) + b(x - c)| ≤ e|x - c|. Thus, for d = min{d', d''}, if |x - c| < d then
|f(x) - f(c)| ≤ e|x - c| and |f'(x) + b(x - c)| ≤ e|x - c|.
Of course, what I need is f(x) ≤ f(c). I don't see how to get that from the above. Any tips?