I had a quick question, If one is given with a bijective function then is is possible to show that the function is monotonic. This question was inspired by a problem which I was recently doing. In the problem this assumption worked out but I still don't know how to prove this fact(if it is one). I know that if the function is bijective then it will have no turning point within its domain as this would create a many to one situation. So its F(x) has to be either positive or negative. The question then is, will the first derivative always bear the same sign? Thanks.

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    $\begingroup$ Bijectivity does not imply continuity or differentiability, but if a function is continuous and bijective, it is monotonic, too. It is a simple consequence of the mean value theorem. $\endgroup$ – Jack D'Aurizio Oct 24 '15 at 14:08
  • $\begingroup$ Could you please elaborate $\endgroup$ – nls Oct 24 '15 at 18:59

Bijectivity does not imply continuity or differentiability.

For instance, $f:(0,1]\mapsto (0,1]$ defined by: $$ f(x)=\left\{\begin{array}{rcl} x &\text{if}& x\in\left(0,\frac{1}{2}\right)\\ \frac{3}{2}-x&\text{if}& x\in\left[\frac{1}{2},1\right]\end{array}\right.$$ is bijective but discountinuous. On the other hand, a function $f:I\mapsto J$ that is continuous and bijective is necessarily monotonic. Proof (by contradiction): assume that $a,b,c\in I$ and $a<b<c$. Assume that $f(b)<f(a)<f(c)$. By the mean value theorem, $f$ attains over $[b,c]$ every value in $[f(b),f(c)]$, in particular $f(a)$, so $f$ cannot be bijective.


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