What is the extreme f(x,y)=xy^2-x+1 ?

The extreme f(x,y)=xy^2-x+1 is Saddle(0,-1),Saddle(0,1)

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extreme f(x,y)=xy^2-x+1

extreme points

f (x, y) = x y^{2} - x + 1

Saddle (0, - 1), Saddle (0, 1)

Solution steps

Find the critical points:

(0, - 1), (0, 1)

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

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

Check the sign of at each critical point

Check critical point

(0, - 1) :

Saddle

Check critical point

(0, 1) :

Saddle

Saddle (0, - 1), Saddle (0, 1)

f (x) = \frac{5}{3} x^{3} \frac{- 65}{2} x^{2} + 200 x - 1

e^{x} (21 + 9 x - x^{2})

f (x) = x^{2} + y^{2} + x^{2} y + 6

f (x) = x^{\frac{4}{5}} (8 x + 32)^{2}

f^{9}

What is the extreme f(x,y)=xy^2-x+1 ?
The extreme f(x,y)=xy^2-x+1 is Saddle(0,-1),Saddle(0,1)