The problem statement is as follows:
Let $y = f(x)$ be a continuous function defined on the closed interval $[0, b]$ with the property that $0 < f(x) < b$ for all $x \in [0, b]$. Show that there exists a point $c \in (0, b)$ with the property that $f(c) = c$. Hint: Consider the function $g(x) = f(x) - x$.
I have wrapping my head around this problem since a couple of days in my free time. Initially it seemed easy. Proving that for some $x \in (0, b)$, $f(x) = c$, is a no brainer. But actually proving that $f(c) = c$ has baffled me. So I thought pitching this question here.
Thanks in advance.
Given $0 < f(x) < b $.
Notice that if $g(x) = x - f(x)$ then
$$ g(0) = - f(0) < 0 $$
while
$$ g(b) = b - f(b) > 0 \; \; \; \; ( since \; \; \; \; f(x) < b \; \; \forall x \in [0,b])$$
Now, use the IVT to conclude!