Prove that if $f(x) = \int_{0}^x f(t)\,dt$, then $f(x) = 0$ 
Prove that if $f(x) = \displaystyle\int_{0}^x f(t)\, dt$ for all $x$, then $f(x) = 0$.

I first differentiated to get $f'(x) = f(x) - f(0)$. Then by the mean value theorem there exists a $c$ in $(0,x)$ such that $f'(c)=\dfrac{f(x)-f(0)}{x}$. Thus, $f'(x) = xf'(c)$. What do I do from here?
A summary of the deliberation in the comments about the necessity of assumptions on $f$: 
There was worry that we needed assumptions on $f$ such as continuity in order to exploit FTC.

Thanks to Aloizio and Clark for pointing out that no assumptions need be placed on $f$ as the integral of a Riemann integrable function is continuous. This gives us that $f$ is continuous (since by assumption we have that $f$ is integrable) and thus the Fundamental Theorem of Calculus applies.


As noted elsewhere, if $f(x)=\int_0^x f(t)\,dt$ for $x\ge 0$, then $f$ is necessarily continuous. 
Being continuous, there is a point $x_0\in[0,1/2]$ such that $f(x_0)\ge f(x)$ for all $x\in[0,1/2]$.  Notice that $f(x_0)\ge f(0)=0$. We then have
$$
f(x_0)=\int_0^{x_0}f(t)\,dt\le\int_0^{x_0}f(x_0)\,dt=x_0f(x_0).
$$
Thus, either $f(x_0)=0$ or   $1\le x_0$, which is absurd. We conclude that $f(x_0)=0$, and so $f(x)\le 0$ for all $x\in[0,1/2]$. In the same way $f(x)\ge 0$ for all $x\in[0,1/2]$. That is, $f$ vanishes on $[0,1/2]$. Repeat this argument on $[1/2,1], [1,3/2],\ldots$ to conclude that $f$ is identically $0$.
