What is the critical f(x,y)=1+(x-y)^4+y^2 ?

The critical f(x,y)=1+(x-y)^4+y^2 is (0,0)

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critical f(x,y)=1+(x-y)^4+y^2

critical points

f (x, y) = 1 + (x - y)^{4} + y^{2}

(0, 0)

Solution steps

1 + (x - y)^{4} + y^{2}

Find where

\nabla f (x, y)

is equal to zero or undefined

(0, 0)

f (x) = x^{2} - 32 x

f (x, y) = x y + \frac{2}{x} + \frac{4}{y}

x^{2} e^{- x^{2}}

2 x^{2} + x y^{2} - 6 x y + 5 x + 2

x^{2} + 4 x + 3

What is the critical f(x,y)=1+(x-y)^4+y^2 ?
The critical f(x,y)=1+(x-y)^4+y^2 is (0,0)