Find the extremum of function f (x, y) = x & sup3; - 4x & sup2; + 2xy-y & sup2; + 5

Find the extremum of function f (x, y) = x & sup3; - 4x & sup2; + 2xy-y & sup2; + 5

f'x=3x^2-8x+2y=0
F'y = 2x-2y = 0--x = y, substituting it into the above formula, we can get 3x ^ 2-6x = 0,
That is, x = 0 or 2, y = 0 or 2
A=f"xx=6x-8
C=f"yy=-2
B=f"xy=2
When x = 0, y = 0, a = - 2, B = 2, C = - 2, AC-B ^ 2 = - 8 * (- 2) - 4 = 12 > 0, a

Why are there three extreme points of function f (x) = (x-1) (x + 1) &# 178; (X-2) &# 179?

That is to find the zero point of its derivative
y'=(x+1)^2(x-2)^3+(x-1)*[2(x+1)(x-2)^3-(x+1)^2*3(x-2)^2]
=(x+1)^2(x-2)^3+2(x-1)(x+1)(x-2)^3-3(x-1)(x+1)^2(x-2)^2
=(x+1)^2(x-2)^3+(x+1)(x-2)^2[2(x-1)(x-2)-3(x-1)(x+1)]
=(x+1)^2(x-2)^3+(x+1)(x-2)^2(-x^2-6x+13)
=(x+1)(x-2)^2(x+1-x^2-6x+13)
=(x+1)(x-2)^2(-x^2-5x+14).
=-(x+1)(x-2)^2(x-2)(x+7).
=-(x+1)(x-2)^3(x+7).
Let y '= 0, we can get three values of X, so there are three extreme points

Extremum of function y = x & # 179; - 27x
What to do after finding the value of X

y'=3x²-27
There are extremums when x = 3 or - 3
y''=6x
When x = - 3, y has a maximum of 54
When x = 3, y has a minimum of - 54