inverse of f(x)=1+1/(x-3) — step-by-step solution

1) For a function

\frac{a x + b}{c x + d}

, two inputs u and v with the same output satisfy

(a d - b \cdot c) \cdot (u - v) = 0

after multiplying by the denominators. Here,

- 3 + 2

-1

≠ 0, so u = v. The function therefore has an inverse on its range

The function domain is

x \in ℝ \ {3}

Express input

x

in terms of output

y

y = \frac{x - 2}{x - 3}

⇥1.1) Let's solve in

ℝ

\frac{x - 2}{x - 3} = y

⇥Let's add fractions:

\frac{x - 2}{x - 3} - y = 0

⇥

\frac{(x - 2) - y \cdot (x - 3)}{x - 3} = 0

⇥

\frac{x - 2 - y \cdot (x - 3)}{x - 3} = 0

⇥1.2) Let's solve in

ℝ

x - 2 - y \cdot (x - 3) = 0

⇥

x - 2 - (y x + y \cdot (-3)) = 0

⇥

x

- 2

- y \cdot x

- y \cdot (-3) = 0

⇥

x

- 2

- y x

+ y \cdot 3 = 0

⇥

x - y x = 2 - y \cdot 3

⇥

(1 - y) \cdot x = 2 - y \cdot 3

⇥Solution obtained:

x = \frac{2 - 3 y}{1 - y}

⇥1.3) To avoid a zero denominator, let's solve:

x - 3 \neq 0

⇥Domain condition:

x \neq 3

Renaming the output as

x

, the inverse is

\frac{2 - 3 x}{1 - x}

, with domain

x \in ℝ \ {1}

▶Inverse function:

\frac{2 - 3 x}{1 - x}