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Range calculation help

  1. Oct 12, 2007 #1
    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

    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. jcsd
  3. Oct 12, 2007 #2
    Interestingly enough [tex]y=\sqrt{25-(x-2)^2}[/tex]

    Is the top half of the circle:


    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 [tex]\sqrt{x}[/tex] belongs to all reals when x >= 0.

    So just find where [tex]25-(x-2)^2=0[/tex] 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
  4. Oct 12, 2007 #3


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    Homework Helper

    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:


    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.


    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.
  5. Oct 12, 2007 #4


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    Homework Helper

    :redface: I didn't notice that it was a circle. So the sketch is straightforward.
  6. Oct 12, 2007 #5
    alright thanks i get it
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