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Show that $$ f(x)=\frac{1}2\sin(2x) + x $$

is invertible.

How do I continue with this?

I've tried with taking the derivative and taken the fact that:

$$ f'(x)=\cos(2x)+1 \geq 0 $$

Is this enough? I'm not sure since it can also equal $0$...

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If you show that it equals $0$ only for discrete values of $x$, then that should be enough. –  user39572 Oct 2 '12 at 16:47
    
Hm, why would that prove the invertible part? To show invertibility, I need to prove that the function is injective. In this case I do it by taking the derivative and try to show that the function is increasing. I cannot relate why discrete values of x that equals 0 would prove that part. –  Curtain Oct 2 '12 at 16:56
    
Because if the derivative is $0$ for discrete values only, then $f$ is never monotone, which implies it is in fact strictly increasing. Therefore it is invertible. –  user39572 Oct 2 '12 at 16:58
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1 Answer 1

up vote 2 down vote accepted

Depends how fussy you are. Suppose that $a\lt b$. We want to show that $f(a)\lt f(b)$.

Step a tiny amount to the right of $a$, say to $c$, where $c\lt b$ and there is no $x$ strictly between $a$ and $c$ such that $f'(x)=0$. Then $f(a)\lt f(c)$. Also, $f(c)\le f(b)$, so $f(a)\lt f(b)$.

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Yeah, that seems logical. It is also easy to explain when we have a trigonometric function to deal with. :-) –  Curtain Oct 2 '12 at 17:03
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@JulianAssange: You are being careful in worrying about the fact that $f'(x)$ can be $0$. The proof I gave in the post can be adapted to show that there is no need to worry unless $f'(x)=0$ in an interval. –  André Nicolas Oct 2 '12 at 17:09
    
Okay, thanks Andre! –  Curtain Oct 2 '12 at 17:10
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