Lowest possible value of a function with derivative greater than 2 I have the following two problems, and want to attempt to solve them with Mathematical rigour(which I don't yet possess):

Suppose that $f$ is differentiable on $[1,4]$ and is such that $f(1) = 10$ and $f' \geq 2$ on $[0,4]$. Find the lowest possible value of $f(4)$

Logic: $f(1) = 10$ and $f' \geq 2$, so we have an increase of minimum $6$ hence, $f(4)\geq 16$
Rigour: $f(1) = 10$ and $f' \geq 2$, so  to have minimum increase, we have the take the lowest possible $f'$, hence $f' = 2$ and $f(x) = 2x + 8$, so $f(4) = 16$.
Not sure if this is even close to rigourous.

Give a function $f$ which is differentiable on $\mathbb{R}$ and which has the following three properties, or explain why such a function cannot exist: $f(0) = -1, f(2) = 4$ and $f'(x) \leq 2$ for all $x$

Logic: It isn't possible as we are increasing by $5y$ in $2x$ which has a greater derivative than 2, thus breaking property three.
Rigour: $f(0) = -1$ has $y = kx+c$ at $(0,-1)$ we have $-1=c$ and with $(2,4)$ we obtain $4 = 2k - 1$, $k = 2.5$ and therefore $f(x) = 2.5x - 1$ and thus $f'(x) = 2.5$, since $2.5 \not\leq 2$, no such function exists.
Could either of these answers be accepted as rigorous?   
 A: For a rigorous answer you need to use Mean Value theorem. By MVT there exists $c \in [1,4]$ such that
$$\frac{f(4)-f(1)}{4-1}=f^{'}(c).$$
But we are given that $f^{'}(c) \geq 2$, therefore
$$f(4)-f(1) \geq 6.$$
Thus the minimum value of $f(4)$ is $f(1)+6.$
A: If you want to be rigorous, apply the fundamental theorem of calculus: 
$f(4) = f(1) + \displaystyle\int_1^4 f'(x)\,dx \ge 10 + \int_1^4 2\,dx = 16$. 
Equality holds iff $f'(x) = 2$ for all $x \in [1,4]$ and $f(1) = 10$. 
A: $f$ differentiable and $f' \geq 2$ on $[0,4]$ gives $f(4)-f(1) = \int_1^4 f'(t) \mathrm{d}t \geq \int_1^4 2 \mathrm{d}t = 6$.  Thus, $f(4) \geq f(1)+6 = 10+6=16$.  This is direct application of the fundamental theorem of calculus.
Use the same method for your second question.
A: For the first bit 
BY mean value theorem, $$\frac{f(4)-f(1)}{3}=f'(\alpha)$$ for some $\alpha \in (1,4)$
Now since $f'(x) \gt 2$ for all $x $ on $[0,4]$, we have $$f(4)-f(1) \ge 6 \implies f(4) \ge 6+10=16 $$
For the second bit similraly, $$\frac{f(2)-f(0)}{2}=f'(\alpha)$$, for some $\alpha \in (0,2)$
Then Since $f'(x) \le 2$ for all $x$ , we have $5 \le 4$, which is not possible
