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Math Help - inverse funtions

  1. #1
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    inverse funtions

    Determine if:
    f(x) = 1/ x-2 and g(x)= 2x +1/x are inverse functions by computing their composistions.
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  2. #2
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    Quote Originally Posted by robinpatrick View Post
    Determine if:
    f(x) = 1/ x-2 and g(x)= 2x +1/x are inverse functions by computing their composistions.
    Hi,

    you get the inverse function if you change the variables.

    If f:y=\frac{1}{x-2} then the inverse funtion of f
    is f^{-1}:x=\frac{1}{y-2}
    (y-2)\cdot x=1
    y-2=\frac{1}{x}
    y=\frac{1}{x}+2
    y=\frac{1}{x}+\frac{2x}{x}
    y=\frac{1+2x}{x} and that's your function g. So g = f^{-1}

    Now proof that the inverse function of g is actually the function f. I leave this to you.

    EB
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  3. #3
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    Quote Originally Posted by robinpatrick View Post
    Determine if:
    f(x) = 1/ x-2 and g(x)= 2x +1/x are inverse functions by computing their composistions.
    Please use parenthesis! f(x) = 1/(x-2) not f(x) = (1/x) - 2 and g(x) = (2x + 1)/x, not g(x) = 2x + (1/x).

    f(g(x)) = f \left ( \frac{2x+1}{x} \right ) = \frac{1}{\left ( \frac{2x+1}{x} \right ) - 2} Multiply top and bottom by x.

    = \frac{1}{\left ( \frac{2x+1}{x} \right ) - 2} \cdot \frac{x}{x} = \frac{1 \cdot x}{\left ( \frac{2x+1}{x} \right )x - 2x} = \frac{x}{2x+1 - 2x}

    = \frac{x}{1} = x (Check!)

    Again:
    g(f(x)) = g \left ( \frac{1}{x-2} \right ) = \frac{2 \left ( \frac{1}{x-2} \right ) + 1}{\left ( \frac{1}{x-2} \right )} Multiply top and bottom by x - 2.

    = \frac{2 \left ( \frac{1}{x-2} \right ) + 1}{\left ( \frac{1}{x-2} \right )} \cdot \frac{x-2}{x-2} = \frac{2 \left ( \frac{1}{x-2} \right ) (x-2) + (x-2)}{\left ( \frac{1}{x-2} \right ) (x-2)}

    = \frac{2 + (x - 2)}{1} = 2 + x - 2 = x (Check!)

    So f(g(x)) = x and g(f(x)) = x. Thus f and g are inverses.

    -Dan
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