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High School
Geometry
Trigonometry
Trigonometric Functions

Carol Gates

Answered question

2021-09-07

Find the inverse the function and the domain and the range of f and $f^{- 1}$ , if the function is $f (x) = \frac{2 x}{x + 5}$ and it's one-to-one.

Answer & Explanation

Cristiano Sears

Skilled2021-09-08Added 96 answers

To find the inverse function, substitute $x = f^{- 1} and y = x$ and calculate:
$y = \frac{2 x}{x + 5}$
$x y + 5 y = 2 x$
$x (y - 2) = - 5 y$
$x = \frac{- 5 y}{2 - y}$
$x = \frac{5 y}{2 - y}$
$f^{- 1} (x) = \frac{5 x}{2 - x}$
If $f (f^{- 1} (x)) = f^{- 1} (f (x)),$
then $f^{- 1} (x)$ is an inverse function of $f (x)$
$= f (f^{- 1} (x))$
$= f (\frac{5 x}{2 - y})$
$= \frac{2 \times \frac{5 x}{2 - y}}{\frac{5 x}{2 - x} + 5}$
$= \frac{\frac{10 x}{2 - x}}{\frac{5 x + 10 - 5 x}{2 - x}}$
=x
The domain of $f (x)$ is range for $f^{- 1} (x)$
And, the range of $f (x)$ is domain for $f^{- 1} (x)$
$f (x) = \frac{2 x}{x + 5}$
$x + 5 = 0$
$x = - 5$
Domain $= (- \infty, - 5) \cup (- 5, \infty)$
$f^{- 1} (x) = \frac{5 x}{2 - x}$
$2 - x = 0$
$x = 2$
Range $= (- \infty, 2) \cup (2, \infty)$

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