Largest possible value of $f(4)$

First, note that since $f$ is continuously differentiable on $\mathbb{R}$ we have that $f(x)f'(x)$ is continuous on $\mathbb{R}$ and so must be bounded on any closed interval $[a,b] \subset \mathbb{R}$ . This implies that $f'(x)f(x)$ is integrable on $[a, b]$ , therefore we can integrate both sides of the inequality $f(x)f'(x) \leq 2$ . We proceed as follows.

Integrating both sides of $f(x)f'(x) \leq 2$ with respect to $x$ over the interval $[0, 4]$ : $\begin{aligned} \int_0^4 f(x)f'(x) dx &\leq \int_0^4 2 dx \end{aligned}$ Applying the fundamental theorem of calculus: $\begin{aligned} \frac{1}{2} \big[f(x)\big]^2\ \bigg|_0^4 &\leq 2x\ \bigg|_0^4 \end{aligned}$ Simplifying: $\begin{aligned} \big[f(4)\big]^2 - \big[f(0)\big]^2 &\leq 16 \end{aligned}$ Using the fact that $f(0)=0$ and simplifying: $\begin{aligned} f(4) &\leq \boxed{4} \end{aligned}$

After solving the differential equation, we get that
maximum f(x) = 2(x^(1/2))