Originally Posted by

**messianic** 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)

when:

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?