Domain/Range of f(x.y)???

Question

for f(x,y) = 3 + sin(SQRT y-e^x)

a) determine and sketch the domain of f in xy plane

b) Determine range for f

*note : it's the square root of y-e^x, so sin of square root of y-e^x

Answer 1

The domain is wherever that square root is defined; i.e., where y-e^x >=0.

So draw y = e^x and shade the area above it (including the curve itself).

Usually the range of sin is everything in [-1,1]. This case is no exception; sin covers everything in that range when (x,y) varies along the line segment, say, from (0,100) to (0,100+ 2 pi).

So the range of f is [2,4]

Answer 2

The range is simple, since -1 <= sin(any value) <= 1, and we are adding this value to 3, we know your possible range is in the closed interval [2, 4].

For your domain, we know our the value we take our square root of needs to be positive (assuming you are not working with complex numbers). We have e^x which we know that the exponential function is fully continuous and positive for all real numbers, so no domain restriction here. I'm not familiar with funtions in two variables, but it would seem that your y value could actually be negative, as long as y > -e^x. So maybe you know how to state the domain in a better fashion, but I would say the domain is x element of (-∞, ∞), and y element of [-e^x, ∞).

I am not familiar with how to graph such a function, as it seems it would require an x,y, and z axis.

--charlie