Inconclusive second partial derivative test Consider the function $$f(x,y)=(x+y)^4$$ and determine whether $f$ has a maximum, a minimum or neither at the point $(0,0)$.
I thought I needed to use the second partial derivative test but how would I go about showing the point is neither a minimum nor a maximum if the test is inconclusive?
 A: Hint: what can you say about the fourth power of any real number?
A: $$f(0, 0) = 0$$
What other values can $f$ assume?  For example, at $(1,1)$, $(-1, 1)$, etc.
A: The Second Derivative Test in single-variable calculus and its analogue for multivariate functions, the second partial-derivative index or Hessian determinant, is of limited help for such functions built on sums of terms using power-functions with exponents larger than $ \ 2 \ . $
One problem we have with the function $ \ f(x,y) \  = \ (x+y)^4 \ $ is that its surface is a sort of "flat-bottomed trench", in which the origin lies on the line of symmetry $ \ y \  = \ -x \ $ , with $ \ f(x,-x) \ = \ 0 \ . $  There is not a critical point, but a line in the plane on which the second derivative of the function equals zero.
The other issue arises if we rotate the coordinate system in the plane by $ \ \frac{\pi}{4} \ $ , using $ \ u \ = \ x  +  y \ $ to write the function as $ \ f(u,0) \ = \ u^4 \ , \ f(0,v) \ = \ 0 \ \ . $  We already know from single-variable calculus that the second derivative will be $ \ 12u^2 \ , $ which is zero at $ \ u = 0 \ $ and positive otherwise (the curve is always "concave upwards").  It is the second-derivative function that tells us about the behavior of the function, rather than simply the value of the second derivative at the single point.
Here, the Hessian determinant is $ \ 12(x+y)^2  ·  12(x+y)^2 \  -  \ [12(x+y)^2]^2 \ = \ 0 \ , $ which is rather unhelpful.  We have at the origin what some authors call a degenerate critical point.  In such situations, we must further investigate the properties of the function and its first derivative in the hope of better understanding these points.
