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Math Help - Inverse functions

  1. #1
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    Inverse functions

    The function f(x)= px+q such that f-1(6)=3 and f-1(-29)= -2. Find f-1(27).

    Thanks.
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Oasis1993 View Post
    The function f(x)= px+q such that f-1(6)=3 and f-1(-29)= -2. Find f-1(27).

    Thanks.
    y = px + q

    x = \frac{1}{p}(y-q). Therefore f^{-1}(x) = \frac{1}{p}(x-q)

    You know that:

    f^{-1}(6) = \frac{1}{p}(6-q) = 3

    f^{-1}(-29) = \frac{1}{p}(-29-q) = -2

    Solve for p and q then sub those values to find f^{-1}(27)
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  3. #3
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    Could you please explain the steps to reach the answer

    thanks.
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  4. #4
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by e^(i*pi) View Post
    y = px + q

    x = \frac{1}{p}(y-q). Therefore f^{-1}(x) = \frac{1}{p}(x-q)

    You know that:

    f^{-1}(6) = \frac{1}{p}(6-q) = 3

    f^{-1}(-29) = \frac{1}{p}(-29-q) = -2

    Solve for p and q then sub those values to find f^{-1}(27)
    \frac{1}{p}(6-q) = 3 ...............(eq1)

    \frac{1}{p}(-29-q) = -2.............(eq2)

    Rearrange eq2 to make q the subject:

    Spoiler:
    -29-q = -2p

    q = -29+2p


    q = -29+2p.........................(eq 2.1)


    Sub the RHS of (eq 2.1) wherever q appears in (eq1):

    \frac{1}{p}(6-(-29+2p)) = 3

    (35-2p) = 3p

    p = 7


    Sub the value of p into (eq2):

    \frac{1}{7}(-29-q) = -2

    -(29+q) = -14

    q = -15

    Test these values

    f^{-1}(6) = \frac{1}{7}(6-(-15)) = 3 . This is correct

    f^{-1}(-29) = \frac{1}{7}(-29-(-15)) = -2. This is correct


    Your equation now shows

    f^{-1}(x) = \frac{1}{7}(x+15) so you should be able to find f^{-1}(27) by direct substitution.

    Spoiler:
    f^{-1}(27) = \frac{1}{7}(27+15) = \frac{42}{7} = 6
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  5. #5
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    Thank you very much for the help!
    it helped alot!!!!!!!
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