$f''(x)\ge 0$, $f'(0)=1$ and $f(x)\le 100$. Does such a function exist? $f''(x)\ge 0$, $f'(0)=1$ and $f(x)\le 100$. Does such a function exist?
I can show that $100-e^{-x}$ does not satisfy the given condition. But I have to show that no function can satisfy the initial condition. 
I can also say that since $f'$ is increasing, using Lagrange's MVT,
$$f'(x)-f(0)\ge1$$

Suppose $f(x) \leq 100$ for all $x$. Now $f'(0) = 1$ and $f''(x) \geq 0$ for all $x$ hence $f'$ is increasing and therefore $f'(x) \geq f'(0) = 1$ for all $x \geq 0$. Hence $f$ is strictly increasing on $[0, \infty)$ and since $f(x) \leq 100$ it follows that $\lim_{x \to \infty}f(x) = L$ exists.
By Mean Value Theorem we can see that $$f(x + 1) - f(x) = f'(\xi)$$ for some $\xi \in (x, x + 1)$. Now we have an obvious contradiction because the LHS tends to $L - L = 0$ as $x \to \infty$ and the RHS is greater than or equal to $1$. Hence there is no such function $f$ with the desired properties.
