Question

решите неравенство f'(x)> 0 :f(x)=1/3x3+4x2-9x+1

Answer 1

$displaystyle f(x)=frac{1}3x^3+4x^2-9x+1\\f'(x)=frac{1}3cdot 3 x^2+4cdot 2x-9 textgreater 0\\x^2+8x-9 textgreater 0\\D=8^2+4cdot 9=64+36=100\\x_1=frac{-8+10}2=frac{2}2=1\\x_2=frac{-8-10}2=-frac{18}2=-9\\(x-1)(x+9) textgreater 0\\underline{quadquad + quadquad-9 quad quad - quad quad 1 quad quad + quad quad}\\\text{OTBET:}quad boxed{xin(-infty;-9)cup(1;+infty)}$