What is the extreme f(x)=x^3-3x+y^2-4y ?

The extreme f(x)=x^3-3x+y^2-4y is Saddle(-1,2),Minimum(1,2)

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extreme f(x)=x^3-3x+y^2-4y

extreme points

f (x) = x^{3} - 3 x + y^{2} - 4 y

Saddle (- 1, 2), Minimum (1, 2)

Solution steps

Find the critical points:

(- 1, 2), (1, 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 2

D (x, y) = 12 x

Check the sign of at each critical point

Check critical point

(- 1, 2) :

Saddle

Check critical point

(1, 2) :

Minimum

Saddle (- 1, 2), Minimum (1, 2)

f (x) = (x^{2} + 13)^{\frac{1}{3}}

f (x) = \frac{x}{x - 2}, 4 \leq x \leq 6

\frac{x}{x ^{2} - 16}

f (x) = 4 sin (2 x)

f (x) = \frac{( 2 x ^{2} + x - 3 )}{( 3 x - 9 )}

What is the extreme f(x)=x^3-3x+y^2-4y ?
The extreme f(x)=x^3-3x+y^2-4y is Saddle(-1,2),Minimum(1,2)