f(x,y) = e ^ (-x^2-y^2)
at point (0,0), D = 0. Therefore the normal min/max test fails.
The problem asks to classify the critical point at the origin by visualizing the surface z = f (x,y)
f(x,0) = e^(-x^2) which is 1 / e(x^2) which would be POSITIVE
f(-x,o) = POSITIVE
f(0,y) = POSITIVE
f(o,-y) = POSITIVE
Shouldn't this mean this is a local/global minimum since as we move away from the point the function increases?
However, the book says (0,0) would be a global maximum. Can someone explain?