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 $0$...
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 $0$...
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 $f(a)\lt f(b)$.