What is the extreme f(x,y)= 3/4 y^2+1/24 y^3-1/32 y^4-x^2 ?

The extreme f(x,y)= 3/4 y^2+1/24 y^3-1/32 y^4-x^2 is Saddle(0,0),Maximum(0,-3),Maximum(0,4)

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extreme f(x,y)= 3/4 y^2+1/24 y^3-1/32 y^4-x^2

extreme points

f (x, y) = \frac{3}{4} y^{2} + \frac{1}{24} y^{3} - \frac{1}{32} y^{4} - x^{2}

Saddle (0, 0), Maximum (0, - 3), Maximum (0, 4)

Solution steps

Find the critical points:

(0, 0), (0, - 3), (0, 4)

\frac{\partial ^{2} f}{\partial x ^{2}} = \frac{3}{2} - \frac{y ( 3 y - 2 )}{8}

D (x, y) = \frac{y ( 3 y - 2 )}{4} - 3

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(0, - 3) :

Maximum

Check critical point

(0, 4) :

Maximum

Saddle (0, 0), Maximum (0, - 3), Maximum (0, 4)

f (x, y) = 2 \cdot \frac{( x ^{2} - 2 \cdot x + 3 )}{y \cdot ( y + 2 )}

f (x) = x^{y}

f (x) = 4

2 x^{3} - 3 x^{2} - 36 x + 5

6 x^{2} - 2 x^{3} + 3 y^{2} + 6 x y

What is the extreme f(x,y)= 3/4 y^2+1/24 y^3-1/32 y^4-x^2 ?
The extreme f(x,y)= 3/4 y^2+1/24 y^3-1/32 y^4-x^2 is Saddle(0,0),Maximum(0,-3),Maximum(0,4)