# Range calculation help

1. Oct 12, 2007

### physstudent1

1. The problem statement, all variables and given/known data

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

3. 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 thats 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?

2. Oct 12, 2007

### Feldoh

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: Oct 12, 2007
3. Oct 12, 2007

### learningphysics

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.

4. Oct 12, 2007

### learningphysics

I didn't notice that it was a circle. So the sketch is straightforward.

5. Oct 12, 2007

### physstudent1

alright thanks i get it