f(x) 2x+1 [x]<3
5x-1 [x]>3
f(x+7)

otvet kakoy

f = (2x-1)^5;             f ' = 10(2x-1)^4f=2x³-x²/2 +4;           f ' = 6x²-xf = 2cosx -3tgx;       f ' = - 2sinx -3/cos²x