Given that $\ f\left( x \right)=ax^{2}+bx+c$, find a value for each of $a, b$ and $c$ such that: $f\left( f'\left( x \right) \right)=f'\left( f\left( x \right) \right)$.
What I did:
$$a=b=c=0\; ∴\; f\left( x \right)=0\; and\; f'\left( x \right)=0\; ∴\; f\left( f'\left( x \right) \right)=f'\left( f\left( x \right) \right)$$
Is this a feasible answer?
HINT: we have $$f(x)=ax^2+bx+c\implies f'(x)=2ax+b$$ $$f(f'(x))=a(2ax+b)^2+b(2ax+b)+c$$ & $$f'(f(x))=2a(ax^2+bx+c)+b$$ hence, $$a(2ax+b)^2+b(2ax+b)+c=2a(ax^2+bx+c)+b$$