a Mean Value Theorem Proof

Question

Disclaimer: this is not homework for marks

Question: Does there exist a differentiable function $f$ with $f(0) = 2$, $f(2) = 5$, and $f'(x) ≤ 1$ for all $x ∈ (0,2)$? If not,why not?

My attempt (incomplete because I don't know what to do): $c \in (0,2)$

$f'(c) = \frac{f(b) - f(c)}{b - c}$

$f'(c) = \frac{f(2) - f(0)}{2 - 0}$

$f'(c) = \frac{5-2}{2}$

$f'(c) = \frac{3}{2}$

Answer 1

The mean value theorem states that if $f$ is a differentiable on $[a,b]$, then there exists a $c\in(a,b)$ with $f'(c)={f(b)-f(a)\over b-a}$. Thus, for $f$ in this question, there would need to be a $c\in(0,2)$ with $f'(c)={f(2)-f(0)\over 2-0}={3\over 2}>1$, so there can't exist such a function.