What is the extreme f(x)=x^2y+y^3-48y ?

The extreme f(x)=x^2y+y^3-48y is Minimum(0,4),Maximum(0,-4),Saddle(4sqrt(3),0),Saddle(-4sqrt(3),0)

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extreme f(x)=x^2y+y^3-48y

extreme points

f (x) = x^{2} y + y^{3} - 48 y

Minimum (0, 4), Maximum (0, - 4), Saddle (43, 0), Saddle (- 43, 0)

Solution steps

Find the critical points:

(0, 4), (0, - 4), (43, 0), (- 43, 0)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y

D (x, y) = 12 y^{2} - 4 x^{2}

Check the sign of at each critical point

Check critical point

(0, 4) :

Minimum

Check critical point

(0, - 4) :

Maximum

Check critical point

(43, 0) :

Saddle

Check critical point

(- 43, 0) :

Saddle

Minimum (0, 4), Maximum (0, - 4), Saddle (43, 0), Saddle (- 43, 0)

x^{2} y - 2 x y + 3 y^{3} - 3 y

\frac{1}{1 + x ^{2}}

- x^{4} - x^{3} + x^{2} + 3 x + 2

x - ln (x)

x^{4} + \frac{4}{3} x^{3} - 4 x^{2}

What is the extreme f(x)=x^2y+y^3-48y ?
The extreme f(x)=x^2y+y^3-48y is Minimum(0,4),Maximum(0,-4),Saddle(4sqrt(3),0),Saddle(-4sqrt(3),0)