Question

Найти критические точки функции: F(x) = 2x^3 + 2,5x^2 - x

Answer 1

$f(x) = 2x^3 + 2,5x^2 - x$
$f'(x) = (2x^3 + 2,5x^2 - x)'=3*2x^2+2*2,5x-1=6x^2+5x-1$
$f'(x)=0$
$6x^2+5x-1=0$
$D=5^2-4*6*(-1)=49$
$x_1= \frac{-5+7}{12}= \frac{1}{6}$
$x_2= \frac{-5-7}{12}=-1$