extreme f(x)=x^3+3x^2+3x+1

解答

极值点

f (x) = x^{3} + 3 x^{2} + 3 x + 1

鞍点 (- 1, 0)

求解步骤

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

Find intervals:

递增

: - \infty < x < - 1,

递增

: - 1 < x < \infty

将

x = - 1

代入

x^{3} + 3 x^{2} + 3 x + 1 : 0

鞍点 (- 1, 0)

e x t re m e 16 (x - 1)^{4} - 4 (x - 1)^{2} + 1

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

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

e x t re m e f (x, y) = 6 x - 4 y - x^{2} - 2 y^{2}

e x t re m e f (x, y) = 10 x^{2} y - 5 x^{2} - 4 y^{2} - x^{4} - 2 y^{4}