SOLUTION 15: Consider the graphs of $y=f(x)$ and $y=x$ on the interval $[0, 1]$ with $f(0)>0$ and $f(1)<1$.

Let function $$g(x)= f(x)-x \ \ \ \ and \ choose \ \ \ \ m=0$$
Function $g$ is continuous on the interval $[0, 1]$ since it is the DIFFERENCE of continuous functions. Note that $$g(0) = f(0) - 0 = f(0)>0$$ and $$g(1)=f(1) - 1 < 0$$
i.e., $m=0$ is between $f(0)$ and $f(1)$.

The assumptions of the Intermediate Value Theorem have now been met on the interval $[0, 1]$, so we can conclude that there is some number $c$ in the interval $[0, 1]$ which satisfies $$g(c)=m$$ i.e., $$f(c)-c =0 \ \ \ \ \longrightarrow \ \ \ \ f(c)=c$$ This completes the proof.