How do you find the inverse of $f (x) = \frac{x}{x + 1}$ $f(x) = x/(x+1)$ and is it a function?

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How do you find the inverse of $f (x) = \frac{x}{x + 1}$ $f(x) = x/(x+1)$ and is it a function?

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Answer 1 · 2016-04-09T12:25:16

If $¯ f (x)$ $bar(f)(x)$ is the inverse of $f (x) = \frac{x}{x + 1}$ $f(x)=x/(x+1)$
then
$XXX ¯ f (x) = \frac{x}{1 - x}$ $color(white)("XXX")bar(f)(x)=x/(1-x)$ (which is a function.)

Explanation:

If $¯ f (x)$ $color(red)(bar(f)(x))$ is the inverse of $f (x)$ $color(blue)(f(x))$
then
$XXX f (¯ f (x)) = x$ $color(white)("XXX")color(blue)(f"(")color(red)(bar(f)(x))color(blue)(")") = x$
and
$XXX f (¯ f (x)) = \frac{¯ f (x)}{¯ f (x) + 1}$ $color(white)("XXX")color(blue)(f"(")color(red)(bar(f)(x))color(blue)(")")=(color(red)(bar(f)(x)))/(color(red)(bar(f)(x))+1)$

$XXX ¯ f (x) = x \cdot (¯ f (x) + 1)$ $color(white)("XXX")color(red)(bar(f)(x))=x*(color(red)(bar(f)(x))+1)$

$XXX ¯ f (x) - x \cdot ¯ f (x) = x$ $color(white)("XXX")color(red)(bar(f)(x))-x*color(red)(bar(f)(x)) =x$

$XXX ¯ f (x) \cdot (1 - x) = x$ $color(white)("XXX")color(red)(bar(f)(x))*(1-x)=x$

$XXX ¯ f (x) = \frac{x}{1 - x}$ $color(white)("XXX")color(red)(bar(f)(x)) = x/(1-x)$

Since this expression provides a unique solution for all valid values of $x$ $x$ ; it is a function.

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Explanation:

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