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I came across a question which is as follows :

Is the following statements always correct ?

1) If $f'(x) > 0$ for all $x$ in the domain of the function $f(x)$ then $f(x)$ is always one-one .

My attempt to the question:

According to me the statement should always be correct since there would be no to $x$ for which the corresponding value of the function will be same . However the answer says it's false . Please explain this .

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    $\begingroup$ This might be a "trick" question, in that even if we assume the derivative $f'(x)$ exists for every $x$ in the domain of $f$, the domain need not be a connected subset of $\mathbb{R}$. Come up with a piecewise defined function on $(0,1) \cup (2,3)$ to serve as a counterexample. $\endgroup$ – hardmath Feb 10 '17 at 17:09
  • $\begingroup$ Could you please elaborate $\endgroup$ – Apoorv Jain Feb 10 '17 at 17:10
  • $\begingroup$ See Michael Hardy's Answer, which illustrates what happens when the domain is not connected (not an interval) in the real line. You can use a couple of intervals where the slope is always $1$, but the range values are duplicated. $\endgroup$ – hardmath Feb 10 '17 at 17:12
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No. It is correct if the domain is an interval. But the derivative of $x\mapsto\tan x$ is positive everywhere in its domain (if the domain excludes points whose tangent is $\infty$), but the function is clearly not one-to-one. You have $\tan 0 = \tan\pi=0,$ so it's not one-to-one.

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  • $\begingroup$ It might be better to stick with the terminology "one-one" meaning injective, rather than "one-to-one" (which many authors use to mean both injective and surjective). $\endgroup$ – hardmath Feb 10 '17 at 17:14
  • $\begingroup$ @hardmath : As far as I know "one-to-one" applied to functions has always meant injective and has never meant surjective. The term "one-to-one" correspondence does mean a bijection, but "one-to-one function" means an injective function. $\endgroup$ – Michael Hardy Feb 10 '17 at 17:29
  • $\begingroup$ My point is mainly to stick with the vocabulary used by the OP, absent some error in terminology. For the meaning of "one-one" or sometimes "1-1", see this older Math.SE post. $\endgroup$ – hardmath Feb 11 '17 at 0:12

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