Step by step solution to "Find the inverse the function and the domain..."

vestirme4

Answered question

2021-10-29

Find the inverse the function and the domain and the range of f and

f^{- 1}

, if the function is

f (x) = 2 \frac{x}{x + 5}

and it's one-to-one.

Sally Cresswell

Skilled2021-10-30Added 91 answers

To find the inverse function, substitute

x = f^{- 1} and y = x

and calculate:

y = 2 \frac{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}

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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