Find the number of functions $f: \mathbb {R\to R^+}$ such that $f(f(x))$ = $f'(x)$ .

The answer is 0.

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$f(x)$ $>$ 0,So $f(f(x))>0$ ,So $f'(x)>0$ ,Hence f(x) is an increasing function.Now apply mean value theorem, $f'(c)$ = $\frac{f(a)-f(0)}{a}$ for some c in between (a,0),a(f(f(c)))=f(a)-f(0).Now since f(x) is increasing,it is very easy to conclude that f(0)<f(f(c)).Plug this in the previous equation and u get f(a)<(a+1)f(f(c))...So any contradiction?????What if a=(-5)...think it is almost done....Hope u liked the problem....For getting more such problems go on upvoting me...XD.. :P