extreme f(x)=x^3+6x^2+12x+7

解答

极值点

f (x) = x^{3} + 6 x^{2} + 12 x + 7

鞍点 (- 2, - 1)

求解步骤

f^{'} (x) = 3 x^{2} + 12 x + 12

Find intervals:

递增

: - \infty < x < - 2,

递增

: - 2 < x < \infty

将

x = - 2

代入

x^{3} + 6 x^{2} + 12 x + 7 : - 1

鞍点 (- 2, - 1)

e x t re m e f (p, q) = 3 p^{3} + 4 q^{3} - 2 p^{2} q - 4 p - 5 q

e x t re m e f (x) = (x - 5) e^{- x + 5 - (x - 5)^{2}}

e x t re m e f (x, y) = 2 x^{2} + 2 x y + 2 y^{2} - 6 x

e x t re m e f (x, y) = 3 x^{2} y + y^{3} - 3 x^{2} - 3 y^{2} - 2

e x t re m e f (x, y) = x y (1 + 2 x - 3 y)