Range Calculation Help: Finding the Range of a Square Root Equation

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To find the range of the equation y = sqrt(25 - (x - 2)^2), it is identified as the upper half of a circle centered at (2, 0) with a radius of 5. The correct domain is determined to be [0, 5], as the square root function only produces non-negative values. The range of the function is therefore [0, 5], reflecting the positive outputs of the square root. The discussion highlights the importance of recognizing the function's geometric representation for accurate range calculation. A sketch of the graph is recommended for clarity in understanding the function's behavior.
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Homework Statement




find the range of sqrt(25-(x-2)^2)

Homework Equations





The Attempt at a Solution



I found the inverse
2+sqrt(-x^2+25)

then I found the domain to be [-5,5] and said that's the range of the original equation however when I graph the original equation the ys only range from [0,+5] what did I mess up?
 
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Interestingly enough y=\sqrt{25-(x-2)^2}

Is the top half of the circle:

(x-2)^2+y^2=25

For there you can easily deduce that it's center is (2,0) and it has a radius of 5. D:[-3,7] and the range is easily deducible as well.

Or you could solve this by realizing that \sqrt{x} belongs to all reals when x >= 0.

So just find where 25-(x-2)^2=0 and that would also give you the same answer for the domain. For the range, you need to know the manipulations of the square root function.
 
Last edited:
y = sqrt(25-(x-2)^2)

sqrt only gives the postive root... hence y >= 0. range will be >= 0

when you take the inverse, you need to include this condition... ie the inverse is

2+sqrt(-x^2+25)... where x must be >= 0, on top of the other condition that you find -5<=x<=5

so x>=0 AND -5<=x<=5, means the domain of this function is 0<=x<=5.

Another way to see it is:

2+sqrt(-x^2+25)

is not a one-to-one function. because a value of x and its negative give the same result. But an invertibe needs to be one-to-one.

However,

2+sqrt(-x^2+25) where x>=0 is one-to-one

But I'm not sure this is the best approach to find the range... you may have non-invertible functions whose range you need to calculate... so in these cases you won't be able to take the inverse.

I think a sketch is the best approach.
 
Feldoh said:
Interestingly enough y=\sqrt{25-(x-2)^2}

Is the top half of the circle:

(x-2)^2+y^2=25

For there you can easily deduce that it's center is (2,0) and it has a radius of 5. D:[-3,7]

:redface: I didn't notice that it was a circle. So the sketch is straightforward.
 
alright thanks i get it
 
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