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

[tex]\displaystyle f(x)=\frac{1}3x^3+4x^2-9x+1\\\\f'(x)=\frac{1}3\cdot 3 x^2+4\cdot 2x-9\ \textgreater \ 0\\\\x^2+8x-9\ \textgreater \ 0\\\\D=8^2+4\cdot 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{\quad\quad + \quad\quad-9 \quad \quad - \quad \quad 1 \quad \quad + \quad \quad}\\\\\\\text{OTBET:}\quad \boxed{x\in(-\infty;-9)\cup(1;+\infty)}[/tex]