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Let $f:(a,b) \subset \mathbb{R} \longrightarrow \mathbb{R}$ a differentiable function on $(a,b)$. In view of the Theorem 2 of this article. Can I conlclude that $f$ is strictly decreasing if and only if $f'(x)<0$, for all $ x \in (a,b)$?

Can I conclude results analogous for strictly increasing function?

Guilherme
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    No. $f(x)= -x^3$ is strictly decreasing but $f'(0) = 0$. – copper.hat Jun 09 '20 at 00:46
  • The "if" part is true, by the mean value theorem. Let $x,y \in (a,b)$ with $x < y$. Then by the MVT there exists $c \in (x,y)$ such that $f(y) - f(x) = f'(c)(y-x)$, and the result follows. –  Jun 09 '20 at 00:48
  • You need $f' \le 0$ everywhere and $f' < 0$ on a dense set – GEdgar Jun 09 '20 at 00:48
  • @Bungo "The part if" means "if $f'<0$ then is $f$ strictly decreasing"? – Guilherme Jun 09 '20 at 00:51
  • For a nice counterexample in the "only if" direction which shows that a strictly monotone differentiable function can have $f'(x) = 0$ on a set of positive measure, see this answer –  Jun 09 '20 at 00:51
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    @GuilhermedeLoreno Yes, that's correct. –  Jun 09 '20 at 00:51
  • the title says increasing, but the question body says decreasing – J. W. Tanner Jun 09 '20 at 00:52
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    @J.W.Tanner I fixed it. – Guilherme Jun 09 '20 at 00:53
  • @Bungo and if $f$ is strictly decreasing and $f>0$ then $f'<0$? – Guilherme Jun 09 '20 at 01:01
  • @GuilhermedeLoreno No, that need not be true, e.g. $f(x) = -x^3 + 1$ on $(a,b) = (-1, 1)$ has $f'(0) = 0$. –  Jun 09 '20 at 01:14
  • @Bungo I have the impression that for functions with behavior like $ f:(0,\infty)\subset \mathbb {R} \longrightarrow \mathbb{R} $ given by $ f(x) = \frac{1}{x}$ the part "only if" holds. – Guilherme Jun 09 '20 at 01:35
  • @GuilhermedeLoreno But that's just one function. "Only if" need not hold in general. You can make a small adjustment to $f(x) = 1/x$ to make it have derivative zero at some point while still maintaining differentiability, strict decreasing, and $f > 0$. E.g. replace $1/x$ in some small neighborhood $[a,b]$ with an appropriate polynomial that matches $1/x$ and its first few derivatives at $a$ and $b$, while satisfying the other properties including $f'(x) = 0$ for some point $x \in (a,b)$. –  Jun 09 '20 at 01:51
  • Yes, just one function. But, if other function satisfies the sames proprierties of $f$ given above, then I have the impression that the "only if" holds. – Guilherme Jun 09 '20 at 01:55

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