this is easier to see (for me) as 1 + e^-(x^2+y^2)
clearly, the domain is all reals for both x and y.
since f(x,y) is even, its maximum is at f(0,0) = 2, and e^-(x^2+y^2) goes to 0, so the range is [1,2]
The level curves are circles, since we want f(x,y) = c
e^-(x^2+y^2) = c
e^(x^2+y^2) = c
x^2+y^2 = ln c
for suitably chosen values of c
Find the domain, the range, and describe the level curves for the function f( x, y) = 1 + e^( -x^2 -y^2)
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