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Is function $f(x)$ is one to one if $f'(x) \ge 0$.

Can we say a function $f(x)$ is one-one if $f(x)$ is Continous and $f'(x) \ge 0$. For example $f(x)=x^3$ is one-one since $$f'(x)=3x^2 \ge 0$$ But why most of the books give $f(x)$ is one-one if $f'(x) \gt 0$. Can

Umesh shankar
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  • $f(x)=3$ is not 1 to 1 – Not a grad student Mar 28 '17 at 02:38
  • You have a reasoning fallacy here. The statement "every function differentiable with positive derivative is one-to-one" is true, and can be proven. But what you say is "every function differentiable with non-negative derivative is one-to-one" (which is false), and your argument is "this must be true, because here is an example." You cannot prove a universal statement ("every function satisfying A is B") by giving an example ("there exists a functions with A that is B"). – Clement C. Mar 28 '17 at 02:43

2 Answers2

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Boring example: a constant function has $f'(x)=0$ everywhere, and is clearly not one-to-one.

Chappers
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No. $f'(x) \ge 0$ does not ensure one-one as shown by the examples in other answers and comments. However, $f'(x) > 0$ does. By the Mean Value Theorem, for $h > 0$, $ \ f(x+h) - f(x) = hf'(\eta) > 0$ for some $\eta \in (x,x+h)$. So $f$ is strictly increasing. Therefore $f(x) = f(y) \Rightarrow x =y$ since otherwise either $x <y$ in which case $f(x) < f(y)$ or $x >y$, in which case $f(x) > f(y)$.

Similarly, $f' < 0$ also makes $f$ one-one.

Andre.J
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  • Ok that i understood clearly. But when $f'(x) \gt 0$ decides injectivity of a function, some functions also have derivative zero at some points.what i learned from all the comments above is, a function $f(x)$ is One-one if $f'(x) \gt 0$ but $f'(x)$ can be zero at finite number of points or countably infinite number of points. Does this make any sense? – Umesh shankar Mar 28 '17 at 03:25