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Thread: Find the range

  1. #1
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    Find the range

    Find the range of the following function.

    f(x) = (4+x)/(3-2x)

    Looking for something like this

    y= x^2
    x = +/- sqrt(y)

    y>0
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  2. #2
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    Quote Originally Posted by glover_m View Post
    Find the range of the following function.

    f(x) = (4+x)/(3-2x)

    Looking for something like this

    y= x^2
    x = +/- sqrt(y)

    y>0
    range of $\displaystyle f(x)$ will be the domain of $\displaystyle f^{-1}(x)$

    $\displaystyle y = \frac{4+x}{3-2x}$

    swap variables ...

    $\displaystyle x = \frac{4+y}{3-2y}$

    solve for $\displaystyle y$ ...

    $\displaystyle 3x - 2xy = 4+y$

    $\displaystyle 3x - 4 = y + 2xy$

    $\displaystyle 3x-4 = y(1 + 2x)$

    $\displaystyle \frac{3x-4}{1+2x} = y = f^{-1}(x)$

    domain of the inverse function is all reals except $\displaystyle x = -\frac{1}{2}$

    so ... range of $\displaystyle f(x)$ is all reals except $\displaystyle y = -\frac{1}{2}$
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  3. #3
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    thanks so much
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