iii) \(f(x) = -2x^2 + 4x\)
\( \lim\limits_{h\to0}\frac{-2*(x+h)^2+4*(x+h)-( -2x^2 + 4x)}{h}=\lim\limits_{h\to0}\frac{-2(x^2+2hx+h^2)+4x+4h+2x^2 - 4x}{h} \)=
=\( \lim\limits_{h\to0}\frac{-2x^2-4hx-2h^2+4x+4h+2x^2 - 4x}{h}= \lim\limits_{h\to0}\frac{-4hx-2h^2+4h }{h}= \lim\limits_{h\to0}-4x-2h+4 =-4x-2h+4\)