inverse of f(x)= 1/(x+8) — 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,

0 \cdot 8 - 1

-1

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

The function domain is

x \in ℝ \ {-8}

Express input

x

in terms of output

y

y = \frac{1}{x + 8}

⇥1.1) Let's solve in

ℝ

\frac{1}{x + 8} = y

⇥Let's sum the fractions:

\frac{1}{x + 8} - y = 0

⇥

\frac{1 - y \cdot (x + 8)}{x + 8} = 0

⇥1.2) Let's solve in

ℝ

1 - y \cdot (x + 8) = 0

⇥

1 - (y x + y \cdot 8) = 0

⇥

1 - y \cdot x - y \cdot 8 = 0

⇥

- y x = - 1 + y \cdot 8

⇥

x = \frac{- 1 + y \cdot 8}{- y}

⇥

x = - \frac{- 1 + y \cdot 8}{y}

⇥

x = \frac{1 - y \cdot 8}{y}

⇥Solution obtained:

x = \frac{1 - 8 y}{y}

⇥1.3) To prevent the denominator from being zero, let's solve:

x + 8 \neq 0

⇥Solution rejected:

x \neq -8

Renaming the output as

x

, the inverse is

\frac{1 - 8 x}{x}

, with domain

x \in ℝ \ {0}

▶Inverse function:

\frac{1 - 8 x}{x}