What is the range of the function f(x,y,z)=e^sqrt(z-5x^2-5y^2)?

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SUMMARY

The function f(x,y,z) = e^sqrt(z - 5x^2 - 5y^2) has a domain defined by the inequality z ≥ 5x^2 + 5y^2, ensuring that the expression under the square root is non-negative. The range of the function is determined by analyzing the output of the exponential function, which results in all positive real numbers. Specifically, as z approaches 5x^2 + 5y^2, the function approaches e^0 = 1, and as z increases indefinitely, f(x,y,z) approaches infinity. Therefore, the range of f(x,y,z) is (1, ∞).

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Homework Statement


Let f(x,y,z)=e^sqrt(z-5x^2-5y^2), find the domain and range of this function


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The Attempt at a Solution


ok i know how to get the domain of the function:
you can't take the sqrt of a negative number you set up the inequality
z-5x^2-5y^2>= 0 and solve for z and you find that z>=5x^2+5y^2

my question is about the range I am not sure how to find it. is it like in 2d all possible numbers in the y, but in 3d its all numbers in z??

thank you
 
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Do it in pieces:

What is the desired range of z-5x^2-5y^2?
What then is the range of h=\sqrt{z-5x^2-5y^2}?
That being the domain of e^h here, what range results?
 

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