Suppose $f:[a,b] \to [a,b]$ is continuous and $f''>0$. Use the fundamental theorem of calculus to argue that if $f(x^*) = x^*$ and $f'(x^*) \geq 1$, then $f(x) > x$ for all $ x > x^*$.
My attempt: By the fundamental theorem, we know that \begin{equation}{ f(x) - f(x^*) = \int_{x^*}^xf'(a)da\geq \int_{x^*}^x 1 da = x-x^* }\end{equation} Then since $f(x^*)=x^*$, we have $f(x) \geq x$. However, I don't think this works because we only know $f'(x^*) \geq 1$, not $f'(a)\geq 1$ for all $a \geq x^*$. Also, the inequality should be $>$ not $\geq$. I wonder if I could use the convexity of $f$ to show that $f'(a) \geq 1$ for all $a \geq x^*$, but I couldn't figure out how to do it, or if I'm just way off track.
Use the fundamental theorem twice. We have, by the fundamental theorem $$ f'(x) = f'(x^*) + \int_{x^*}^x f''(\xi)\, d\xi > f'(x^*) \ge 1, \quad x > x^* $$ So $f'(x) > 1$ for $x > x^*$. Now $$ f(x) = f(x^*) + \int_{x^*}^x f'(\xi)\, d\xi > f(x^*) + (x-x^*) = x. $$