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The First Derivative Test says that the function f is continuous on [a,b] and differentiable on (a,b), except possibly at c in (a,b).

a) We can prove that if f ' (x) > 0 for x in (a,c), and f ' (x) < 0 for x in (c,b), then f has a relative maximum at x = c.

Can we prove or disprove the converse of statement a?

Formal and informally.

k_math
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  • The converse is false. It's not so easy to find counterexamples, but they exist. See here: https://math.stackexchange.com/questions/3620613/is-the-reverse-of-the-first-derivative-test-statement-true – Mark Apr 21 '22 at 20:20
  • @Mark is there a formal way to prove that function using f'(a) = lim (x -> a) f(x)-f(a) / x-a – k_math Apr 21 '22 at 20:36

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The opposite needn't to be true; the function $$ f(x) = x^2 (\sin\frac1x - 1), x \ne 0, f(0)=0 $$ has maximum at $0$, however, there is no intervals $(a,0), (0,b)$ such that $f'$ is negative on $(a,0)$ and positive on $(0,b)$.

Darweesh
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