$f''(x)\ge 0$, $f'(0)=1$ 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,
$$f'(x)-f(0)\ge1$$
 A: Assume $x>0$.
We have $f'(x) = f'(0) + \underbrace{\int_0^x f''(t)\,\mathrm dt}_{\ge 0}\ge f'(0) = 1$.
Therefore $f(x) = f(0) + \int_0^x f'(x)\,\mathrm dt \ge f(0) + \int_0^x 1\,\mathrm dt = f(0)+x$.
Thus $f(x) \le 100$ implies $f(0) + x \le 100$, that is $x \le 100-f(0)$. Obviously this inequality will be violated for sufficiently large $x$.${}{}$
A: By the Mean Value Theorem, for any $x>0$ there exists $t\in(0,x)$ such that
$$ f(x)=f(0)+f^{\prime}(t)x$$
and since $f^{\prime}$ is increasing, it follows that
$$ f(x)\geq f(0)+f^{\prime}(0)x=f(0)+x$$
If $x$ is sufficiently large, this will be greater than $100$.
A: No such function exists. I'll give you a more formal but probably excessive proof! Suppose otherwise:
Since $f''(x) \geq 0$, then $f'(x)$ is increasing for all $x \in \mathbb{R}$. Then we have $f'(x) > 1$ for all $x \gt 0$. So $f$ is increasing. Since $f$ is increasing and bounded above by 100, then by the monotone convergence theorem (you should know this from calculus 1), then the sequence $f(1), f(2), f(3), ...$ must converge to an upper limit $S \leq 100$ as $x\to\infty$. 
Since the sequence is convergent, then by definition, for some $N \in \mathbb{N}$, $n > N$ implies $|f(n) - S| < 0.1$, so we have: 
$$\begin{align}|f(N + 2) - f(N+1)| &= |(f(N + 2) - S) - (f(N+1) - S)|\\ &\leq |f(N + 2) - S| + |f(N + 1) - S|\\ &\lt 0.2\end{align}$$
Now we can use the mean value theorem. Since $f$ is continuous, we see that there exists $c \in (N+1, N+2)$ such that $f'(c) = \frac{f(N+2) - f(N+1)}{(N+2) - (N+1)} \lt \frac{0.2}{1} = 0.2$. But this contradicts our knowledge that $f'(x) > 1$ for $x \geq 0$.
Thus, our proof is complete via contradiction.
A: 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.
A: Let $g(x)=f(x)-x-f(0)$.  Then $g'(x)=f'(x)-1$ and $g''(x)=f''(x)$.  The assumption $f''(x)\ge0$ implies $g''(x)\ge0$ which implies $g'(x)\ge g'(0)=f'(0)-1$ for $x\ge0$.  The assumption $f'(0)=1$ now tells us that $g'(x)\ge0$ for $x\ge0$, from which is follows that $g(x)\ge g(0)=f(0)-0-f(0)=0$ for all $x\ge0$.  In particular, 
$$g(101+|f(0)|)=f(101+|f(0)|)-(101+|f(0)|)-f(0)\ge0$$
hence
$$f(101+|f(0)|)\ge101+|f(0)|-f(0)\ge101\gt100$$
