If f(x) = 1+ 1/(x-1), prove that f(x) is its own inverse?

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  • 4 weeks ago
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    Let's find the inverse:

    Using y instead of f(x) for now:

    f(x) = 1 + 1/(x - 1)

    y = 1 + 1/(x - 1)

    Now switch the variables and solve for y again:

    x = 1 + 1/(y - 1)

    Multiply both sides by (y - 1):

    x(y - 1) = 1(y - 1) + 1

    Expand:

    xy - x = y - 1 + 1The right side simplifies:

    xy - x = y

    Move all terms with "y" to the left and all others to the right:

    xy - y = xFactor out the y:

    y(x - 1) = x

    y = x / (x - 1)

    Now to check to see if that's the same, let's simplify the original equation:

    y = 1 + 1 / (x - 1)

    Common denominator:

    y = (x - 1) / (x - 1) + 1 / (x - 1)

    Add the numerators:

    y = (x - 1 + 1) / (x - 1)

    simplify:

    y = x / (x - 1)And both the inverse and the simplified original equation are the same.

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