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Inverse of radical function

  1. Jul 17, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the inverse of each of the following functions


    2. Relevant equations

    y= [sqrt] x^2 + 9
    3. The attempt at a solution

    y = [sqrt] x^2 +9
    x= [sqrt] y^2 +9
    x-3= y

    I did the sqrt of 9, and sqrted y and its wrong.

    The answer is apparently y=+/(plusminus) [sqrt]x^2 -9 .. How do you solve this?
     
  2. jcsd
  3. Jul 17, 2011 #2
    anyone?
     
  4. Jul 17, 2011 #3

    tiny-tim

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    Hi Nelo! :smile:

    (have a square-root: √ and try using the X2 icon just above the Reply box :wink:)
    How did you get that second line? :confused:
     
  5. Jul 17, 2011 #4
    Its an inverse...
     
  6. Jul 17, 2011 #5

    tiny-tim

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    No it isn't. :redface:

    Write it out in full before trying to invert it. :smile:
     
  7. Jul 17, 2011 #6

    eumyang

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    Need some parentheses. I don't know if you mean
    [itex]y = \sqrt{x^2} + 9[/itex]
    or
    [itex]y = \sqrt{x^2 + 9}[/itex]


    Looks like the 9 is INSIDE the square root. You got it completely wrong from line 2 to line 3. From line 2, square both sides, and then subtract the 9. Then solve for y.
     
  8. Jul 18, 2011 #7

    HallsofIvy

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    There are two commonly taught ways to find inverses of functions:
    1) First solve y= f(x) for x, then swap x and y.
    2) First swap x and, y, then solve for y.

    So given [itex]f(x)= \sqrt{x^2+ 9}[/itex], you can write that, first, as [itex]y= \sqrt{x^2+ 9}[/itex] then swap x and y to write [itex]x= \sqrt{y^2+ 9}[/itex].

    However, it is not true that [itex]\sqrt{a^2+ b^2}= a+ b[/itex]. For example, [itex]\sqrt{9+ 16}= \sqrt{25}= 5[/itex], not [itex]\sqrt{9}+ \sqrt{16}= 3+ 4= 7[/itex].

    In order to solve [itex]x= \sqrt{y^2+ 9}[/itex] for y, start by squaring both sides.
     
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