Question

Решите неравенство f'(x)<0, если:
а) f(x)= x^3+1.5x^2
б) f(x)= 4x-1/3x^3

Answer 1

a) f'(x) =3x^2+3x <0

3x*(x+1)<0

x=0 x+1=0

x=-1

(-1;0)

b)f'(x)=4-1/3*3x^2 = 4-x^2<0

x^2-4>0

(x-2)(x+2)>0

x=2 x=-2

(-oo;-2)U(2;+oo)