The given function is f ( x , y ) = 1 x 2 + y 2 − 1 From my understanding, a rational function such as this one would not have any points in which ∇ f ( x , y ) = 0 , so I would conclude that no such points exist. Would this be true or am I not looking at this in a different way?

Question

The given function is  f  (  x  ,  y  )  =      1                  x        2            +              y        2            −      1      From my understanding, a rational function such as this one would not have any points in which   ∇  f  (  x  ,  y  )  =      0  , so I would conclude that no such points exist. Would this be true or am I not looking at this in a different way?

glorietka4b · Accepted Answer

You need to compute the gradient of   f  (  x  ,  y  ):            ∂      f              ∂      x        =  −            2      x              (              x        2            +              y        2            −      1              )        2              ,              ∂      f              ∂      y        =  −            2      y              (              x        2            +              y        2            −      1              )        2            Hence   ∇  f  (  x  ,  y  )  =  0    ⟺    (  x  ,  y  )  =  (  0  ,  0  ). Therefore   (  0  ,  0  ) is the only stationary point of   f.

Aidyn Cox · Accepted Answer

by the quotient rule we get      f    x    =  −            2      x              (              x        2            +              y        2            −      1              )        2                  f    y    =  −            2      y              (              x        2            +              y        2            −      1              )        2            solve the system      f    x    =  0  ,      f    y    =  0

The given function is f ( x , y ) = 1 x

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