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Math Help - Simple Inverse Function problems!

  1. #1
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    Simple Inverse Function problems!

    So the first one is g(x)=(x/2)+1

    Find g^-1 [g(2)]

    Second is h(x)=(x/3)+1
    Find h[h^-1 (x)]
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  2. #2
    Super Member bigwave's Avatar
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    Talking with inverse functions exchange x for y and solve for y

    first note that these functions are one to one and that from:

    g(x) = \frac{x}{2} \ + \ 1

    g(2) = \frac{2}{2} \ + \ 1 \ = 2

    to get the inverse temporarily change this equation to

    y = \frac{x}{2} \ + 1

    now exchange x for y and solve for y

    x = \frac{y}{2} + 1

    y = 2x - 2 \    or   \ g^{-1}(x) = 2x - 2

    then\ g^{-1}[g(2)] = 2(2) - 2 \ = \ 2

    now as for h(x) = \frac{x}{3} + 1

    h^{-1}(x) = 3x -3

    so h[h^{-1}(x)] = \frac{3x-3}{3} - 3 = x-4
    Last edited by mr fantastic; November 5th 2009 at 03:55 AM. Reason: Increased font size.
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  3. #3
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    Quote Originally Posted by Thatoneguy12345 View Post
    So the first one is g(x)=(x/2)+1

    Find g^-1 [g(2)]

    Second is h(x)=(x/3)+1
    Find h[h^-1 (x)]
    You don't need to do all the calculation bigwave did (the first is correct and nicely done- for the second he found h^-1(x), not h[h^-1(x)]) and it really doesn't matter how g and h are defined (as long as they have inverses).

    By the definition of inverse, g^-1[g(2)]= 2 and h[h^-1(x)]= x!


    If f is any function having an inverse and a, b is any numbers in the domains of f and f^-1 respectively, f^-1(f(a))= a and f(f^-1(a))= a. That's the definition of "inverse function".
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