Сritical point ( 0 , 0 ) of the function f ( x , y ) = x 2 + 4 y 3 . Since the Hessian is singular, we cannot conclude anything about the local extrema. However, a graph of this function clearly shows that (0,0) is neither a local max nor min. Show this without just referring to the graph. Further, is there a way to show that ( 0 , 0 ) is isolated or not isolated?

Question

Сritical point   (  0  ,  0  ) of the function   f  (  x  ,  y  )  =      x    2    +  4      y    3  . Since the Hessian is singular, we cannot conclude anything about the local extrema. However, a graph of this function clearly shows that (0,0) is neither a local max nor min. Show this without just referring to the graph. Further, is there a way to show that   (  0  ,  0  ) is isolated or not isolated?

Kennedy Evans · Accepted Answer

A critical point   p is not a local extremum if every neighborhood of   p contains a point   q with   f  (  q  )  &amp;gt;  f  (  p  ) and a point       q    ′   with   f  (      q    ′    )  &amp;lt;  f  (  p  ).In this case,   f  (  0  ,  0  )  =  0, but for any   ϵ  &amp;gt;  0,   f  (  ϵ  ,  0  )  =      ϵ    2    &amp;gt;  0, and   f  (  ϵ  ,  0  )  =      ϵ    2    &amp;gt;  0.

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