If $f (x) = \frac{2 x + 5}{x}$ $f(x) = (2x+5)/x$ , what is $f^{- 1} (x)$ $f^-1(x)$ ?

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If $f (x) = \frac{2 x + 5}{x}$ $f(x) = (2x+5)/x$ , what is $f^{- 1} (x)$ $f^-1(x)$ ?

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Answer 1 · 2017-06-27T06:43:02

$f^{- 1} (x) = \frac{5}{x - 2}$ $f^-1(x)=\frac{5}{x-2}$

Explanation:

$f (x) = \frac{2 x + 5}{x}$ $f(x)=(2x+5)/x$
$x = f^{- 1} (\frac{2 x + 5}{x})$ $x=f^-1((2x+5)/x)$

Let $y = \frac{2 x + 5}{x}$ $y=(2x+5)/x$
Now solve $y = \frac{2 x + 5}{x}$ $y=(2x+5)/x$ for $y$ $y$
$y x = 2 x + 5$ $yx=2x+5$
$y x - 2 x = 5$ $yx-2x=5$
$x (y - 2) = 5$ $x(y-2)=5$
$x = \frac{5}{y - 2}$ $x=5/(y-2)$

So,
$x = f^{- 1} (\frac{2 x + 5}{x})$ $x=f^-1((2x+5)/x)$
$\frac{5}{y - 2} = f^{- 1} (y)$ $5/(y-2)=f^-1(y)$
And
, $f^{- 1} (x) = \frac{5}{x - 2}$ $f^-1(x)=\frac{5}{x-2}$

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If $f (x) = \frac{2 x + 5}{x}$ $f(x) = (2x+5)/x$ , what is $f^{- 1} (x)$ $f^-1(x)$ ?

1 Answer

Explanation:

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