# Explain why the range of function f(x,y)=square root[100-(x^2+y^2)] is [0;10].

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### 2 Answers

The range of a function is the value that the independent variables can take to give a real value for f(x).

Here we have f(x, y) = sqrt [ 100 - (x^2 + y^2)].

For f(x, y) to give real values 100 - (x^2 + y^2) >=0

=> (x^2 + y^2) =< 100

=> x^2 + y^2 =< 10^2

As we can see this is the equation of a circle and (x, y) can take on all values on the circle or within it but neither of them can exceed the value 10 or go lower than 0.

**This proves that the range of the function f(x, y) is [0 , 10]**

Let's determine the range of 2 variables function. First, we'll determine the domain of the function.

The domain of the function has to contain the values of the variables thatsatisfy the function .

In this case, because of the constraint that the radicand has to be positive or at least zero, we'll get the domain of the function:

D = {(x,y) /100-x^2-y^2 >=0}

D = {(x,y) /x^2 + y^2 >= 100}

The domain is represented by the disc whose center is the origin of the coordinates system and the radius is 10.

We'll determine the range:

z = {z/z = sqrt(100-x^2-y^2), (x,y) belongs to D}

Since z>=0 andÂ 100-x^2-y^2 =< 100 => sqrt(100-x^2-y^2)=<10 (the values of the function could not overstep the radius of the disc r = 10). That is why the range of the function is the closed interval [0,10].

**The the range of 2 varaible function is indeed the closed interval [0,10].**