Answer
No function $f$ exists such that $f(0) = -1$, $f(2) = 4$, and $f'(x) \leq 2$ for all $x$.
Work Step by Step
Let's assume that $f'(x) \leq 2$ for all $x$
Since $f$ is differentiable for all $x$ then $f$ is continuous for all $x$. Then $f$ is continuous and differentiable on the interval $[0,2]$
According to the Mean Value Theorem, there is a number $c$ in the interval $(0,2)$ such that $f'(c) = \frac{f(2)-f(0)}{2-0} = \frac{4-(-1)}{2} = \frac{5}{2}$
However, this contradicts our assumption that $f'(x) \leq 2$ for all $x$.
Therefore, no function $f$ exists such that $f(0) = -1$, $f(2) = 4$, and $f'(x) \leq 2$ for all $x$.
