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.
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 $f(2) > 3$.