TITLE: If $f(0)=f'(0)=1$ and $f''(x)>0$, then $f(2)>2$
QUESTION [6 upvotes]: I stumbled upon this problem and I'd like to see if these arguments are correct.  
Given $f(0)=f'(0)=1$ and $f''(x)>0$  for all $x\geq0$, then $f(2)>2$.  Assume $f(x)$ is double differentiable for $x\geq0$. (The problem doesn't state much else so I'm assuming other conditions such as $f$ is a real-valued one variable function are implicit).  
ARGUMENT 1. 
Since $f''(x)>0$, then $f'(x)$ is increasing. Therefore  
$\qquad \qquad f'(x)\geq f'(0)=1$ $\qquad$for all $x\geq 0$.$\quad$ Then 
$$\int_{0}^{2} f'(x) dx \geq \int_{0}^{2} dx$$
$$f(2)-f(0) \geq 2$$
$$f(2) \geq 2+f(0)=3$$
Since $f(2) \geq 3$, it follows that $f(2)$ is strictly greater than 2. 
ARGUMENT 2
By the mean value theorem, it exists a $c\in (0,2)$ such that  
$$\frac{f(2)-f(0)}{2} =f'(c)  \geq1$$
Then 
$$ f(2)-1 \geq 2 $$ 
$$f(2)\geq 3$$
And it concludes the same way. 
Some feedback or notes about specifying things I left implicit or unclear would be welcome. 
NOTE: I intend to apply this exercise in a mid-term test for freshmen on mathematics major, so I'd love to be as clear and specific as I could be.

REPLY [4 votes]: Both arguments are correct. As the first step you used that 
$f''(x) > 0$ implies that $f'(x)$ is increasing, that is a consequence
of the mean-value theorem.
As the second step you used the "second fundamental theorem of calculus":
$$
 f(2) - f(0) = \int_0^2 f'(t) \, dt  \ge (2 - 0)\cdot 1
$$
or the mean-value theorem:
$$
 \frac{f(2)-f(0)}{2-0} =f'(c) \ge 1 \text{ for some $c$ between $0$ and $2$}
$$
The outcome is the same because $f'$ is continuous.
(In general, the second approach is more versatile because it only 
requires $f$ to be differentiable, but not that $f'$ is
integrable.)
If you know the "Taylor formula with remainder" then you can
proceed in one step:
$$
 f(x) = f(0) + f'(x)(x-0) + \frac{f''(c)}{2!} (x-0)^2 
$$
for some $c$ between $0$ and $x$. In your case
$$
 f(x) = 1 + x + \frac{f''(c)}{2!} x^2 > 1 + x
$$
for $x > 0$, and therefore $f(2) > 3$.