A real function problem Let $f : \mathbb{R} \to \mathbb{R}$ be a function with continuous derivative such that $f(\sqrt{2}) = 2$ and $$f(x) = \lim_{t \to 0}\dfrac{1}{2t}\int_{x-t}^{x+t}sf'(s)\,ds \ \text{for all} \ x \in \mathbb{R}.$$ Then $f(3)$ equals 
$(a) \ \ \sqrt{3} \hspace{1.25 in} (b) \ \ 3\sqrt{2} \hspace{1.25 in} (c) \ \ 3\sqrt{3} \hspace{1.25 in} (d) \ \ 9$
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Can someone help me solve this problem on real function .
Here is my try 
I differentiated both sides with respect to $x$. And I got derivative of $x$ to be $0$.
I concluded $f(x)$ is a constant function.
So answer should be $2$. But $2$ is not an option . 
How to solve this ??

We have:
\begin{align}
f(x)&=\lim_{t \rightarrow 0}\frac{1}{2t}\int_{x-t}^{x+t}sf'(s)\,ds\\
&=\lim_{t \rightarrow 0}\frac{1}{2t}sf(s)\,\Big|_{x-t}^{x+t}-\lim_{t \rightarrow 0}\frac{1}{2t} \int_{x-t}^{x+t}f(s)\,ds\\
&=\lim_{t \rightarrow 0}\frac{1}{2t}sf(s)\,\Big|_{x-t}^{x+t}-f(x)
\end{align}
or
\begin{align}
2f(x)&=\lim_{t \rightarrow 0}\frac{1}{2t}s f(s)\,\Big|_{x-t}^{x+t}\\
&=\lim_{t \rightarrow 0}\frac{1}{2t}\cdot \left[ (x+t)f(x+t) - (x-t)f(x-t) \right]\\
&=\lim_{t \rightarrow 0}\frac{1}{2t}\cdot \left[ x(f(x+t) - f(x-t)) +t(f(x+t) + f(x-t)) \right]\\
&=x \lim_{t \rightarrow 0}\frac{f(x+t) - f(x-t)}{2t} + \lim_{t \rightarrow 0} \frac{f(x+t) + f(x-t)}{2}\\
&=xf'(x) + f(x)
\end{align}
Or 
$$f(x)=xf'(x)$$
which is $$f(x)=c x$$
or, from $f(\sqrt{2})=2$,
$$f(x)=\sqrt{2} x$$
the answer is (b)
