I need to prove this: Let $f$ be a strictly monotonic continuous real-valued function defined on $[a,b]$ such that $f(a)<a$ and $f(b)>b$, then $\exists$ exactly one c $\in$ $(a,b)$ such that $f(c)=c$
My attempt
I could figure out from the given data that $f$ is strictly increasing. I tried to construct a new function:
$g(x)=f(x)-x$
and I got $g(a)<0$ and $g(b)>0$. I now know for sure that there exits at least one such c $\in$ $(a,b)$ such that $f(c)=c$. But how do I conclude there is exactly one such $c$?
Please suggest. Thanks in advance.
EDIT I may have chosen the wrong one but there are other options to choose from:
- $\exists$ exactly one c $\in$ $(a,b)$ such that $f(c)=c$
- $\exists$ $c_1, c_2$ $\in$ $(a,b)$ such that $f(c_1)=c_1$ and $f(c_2)=c_2$
- $\exists$ no such c $\in$ $(a,b)$ such that $f(c)=c$
- $\exists$ infinitely many such c $\in$ $(a,b)$ such that $f(c)=c$
NOT TRUE.
Take $$ f(x)=x+\frac{1}{2}\sin x. $$ Then $f'(x)>0$ and $f(x)=x$, whenever $x=k\pi$.