Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
  4. Jul 17, 2011 #3


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor
    Homework Helper

    No it isn't. :redface:

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


    User Avatar
    Homework Helper

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


    User Avatar
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook