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

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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]

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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 and  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].

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