Question

Найти критические точки:
y=x^3+x^2-5x-11

Answer 1

$y=x^3+x^2-5x-11\\y'=3x^2+2x-5$

$3x^2+2x-5=0\\D=\sqrt{2^2-4*3*(-5)} =\sqrt{64} =б8\\\\x_1=\frac{-2+8}{6}=1 \\\\x_2=\frac{-2-8}{6} =-\frac{5}{3}$