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

The extreme f(x)=x^3+x^2y-y^2-4y is Maximum(0,-2),Saddle(1,-3/2),Saddle(-4,6)

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

extreme points

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

Maximum (0, - 2), Saddle (1, - \frac{3}{2}), Saddle (- 4, 6)

Solution steps

Find the critical points:

(0, - 2), (1, - \frac{3}{2}), (- 4, 6)

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

D (x, y) = - 2 (6 x + 2 y) - 4 x^{2}

Check the sign of at each critical point

Check critical point

(0, - 2) :

Maximum

Check critical point

(1, - \frac{3}{2}) :

Saddle

Check critical point

(- 4, 6) :

Saddle

Maximum (0, - 2), Saddle (1, - \frac{3}{2}), Saddle (- 4, 6)

y = x^{\frac{4}{5}} (x - 1)

f (x) = \frac{( x ^{2} - 11 )}{x + 6}

\frac{64}{x ^{2} + 6 x - 7}

f (x) = 6 x e^{- x^{2}}

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

What is the extreme f(x)=x^3+x^2y-y^2-4y ?
The extreme f(x)=x^3+x^2y-y^2-4y is Maximum(0,-2),Saddle(1,-3/2),Saddle(-4,6)