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Thread: Combination of Functions

  1. #1
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    Combination of Functions

    Find the inverse of each function.

    g)

    $\displaystyle y = \frac {x + 5}{x -5}$

    Where did I go wrong?

    $\displaystyle x = \frac {y + 5}{y - 5}$

    $\displaystyle x(y - 5) = y + 5$

    $\displaystyle x(y - 5) -5 = y$

    $\displaystyle xy - 5x - 5 = y$

    $\displaystyle -5x - 5 = y - xy$

    $\displaystyle -5x - 5 = y(1 - x)$

    $\displaystyle \frac {-5x - 5}{1 - x} = y$

    $\displaystyle \frac {-5(x + 1)}{1 - x} = y$

    Even if I move it to the right side. . .I get a different answer. . .

    y = $\displaystyle \frac {5(x - 1)}{x - 1}$

    Correct answer is. . .

    $\displaystyle \frac {5(x + 1)}{x - 1}$


    h)

    $\displaystyle y = \frac {1}{x^2}$

    Is my answer correct?

    $\displaystyle x(y^2) = 1$

    $\displaystyle y^2 = \frac {1}{x}$

    $\displaystyle y = \sqrt{\frac {1}{x}}$
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Macleef View Post
    Find the inverse of each function.

    g)

    $\displaystyle y = \frac {x + 5}{x -5}$

    Where did I go wrong?

    $\displaystyle x = \frac {y + 5}{y - 5}$

    $\displaystyle x(y - 5) = y + 5$

    $\displaystyle x(y - 5) -5 = y$

    $\displaystyle xy - 5x - 5 = y$

    $\displaystyle -5x - 5 = y - xy$

    $\displaystyle -5x - 5 = y(1 - x)$

    $\displaystyle \frac {-5x - 5}{1 - x} = y$

    $\displaystyle \frac {-5(x + 1)}{1 - x} = y$

    Even if I move it to the right side. . .I get a different answer. . .

    y = $\displaystyle \frac {5(x - 1)}{x - 1}$

    Correct answer is. . .

    $\displaystyle \frac {5(x + 1)}{x - 1}$
    You have:

    $\displaystyle \frac {-5(x + 1)}{1 - x} = y$

    or:

    $\displaystyle y=\frac {-5(x + 1)}{1 - x}=\frac{5(x+1)}{(x-1)}$

    (that is multiply top and bottom of the middle term side by $\displaystyle -1$ to get the right most term, which is the desired answer)
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Macleef View Post
    h)

    $\displaystyle y = \frac {1}{x^2}$

    Is my answer correct?

    $\displaystyle x(y^2) = 1$

    $\displaystyle y^2 = \frac {1}{x}$

    $\displaystyle y = \sqrt{\frac {1}{x}}$
    Subject to the restriction that $\displaystyle x > 0$ yes except that you have reversed the meaning of $\displaystyle x$ and $\displaystyle y$.


    $\displaystyle y(x^2) = 1$

    $\displaystyle x^2 = \frac {1}{y}$

    $\displaystyle x = \sqrt{\frac {1}{y}}$

    when $\displaystyle y > 0$


    RonL
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