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Thread: 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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    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.
    $\displaystyle y = px + q$

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

    You know that:

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

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

    Solve for p and q then sub those values to find $\displaystyle f^{-1}(27)$
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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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    Quote Originally Posted by e^(i*pi) View Post
    $\displaystyle y = px + q$

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

    You know that:

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

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

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

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

    Rearrange eq2 to make q the subject:

    Spoiler:
    $\displaystyle -29-q = -2p$

    $\displaystyle q = -29+2p$


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


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

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

    $\displaystyle (35-2p) = 3p$

    $\displaystyle p = 7$


    Sub the value of p into (eq2):

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

    $\displaystyle -(29+q) = -14$

    $\displaystyle q = -15$

    Test these values

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

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


    Your equation now shows

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

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