Solutions

Consider the continuous function $h=f-g$ on $[a,b]$. We have $h(a)=f(a)-g(a)<0$ and $h(b)=f(b)-g(b)>0$. By Theorem 1, there exists $x$ in $[a,b]$ such that $h(x)=0$. But this means that $f(x)=g(x)$.