Solutions

a) Here, we suppose for contradiction that $f(x)\ne -x+1=g(x)$ for all $x$ and employ the same proof outline as in Exercise 11. We have that $f(0)<g(0)=1$ and $f(1)>g(1)=0$. Hence, by Exercise 10, $f(x)=-x+1=g(x)$ for some $x$ in $[0,1]$.

b) Again, we employ a proof by contradiction supposing that $f(x)\ne g(x)$ for all $x$. By applying Exercise 10, we end up with a contradiction for both cases.