Question

Найдите точки экстремума и определите их вид: f(x)=x^3-2x^2-x+5

Answer 1

$y=x^3-2x^2-x+5\y'=3x^2-4x-1\3x^2-4x-1=0\bf x_1=frac{2}3-frac{sqrt7}3quad-quad bf Maximum\x_2=frac{2}3+frac{sqrt7}3quad-quad Minimum$

Answer 2

f(x) = x^3 - 2x^2 - x + 5

f " (x) = (x^3 - 2x^2 - x + 5) "  =  3x^2 - 4x - 1

3x^2 - 4x - 1 = 0

D = b^2 - 4ac = 16 + 12 = 28 > 0

x_1 = (-b - Vd)/2a =  (4 - V28)/6 = (4 - 2V7)/6 = (2 - V7)/3     max

x_2 = (-b + Vd)/2a = (2 + V7)/3       min