inverse of f(x)=(6x+4)/(x-1) — 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,

6 \cdot (-1) - 4

-10

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

Express input

x

in terms of output

y

y = \frac{6 x + 4}{x - 1}

⇥1.1) Let's solve in

ℝ

\frac{6 x + 4}{x - 1} = y

⇥1.2) Let's add fractions:

\frac{6 x + 4}{x - 1} - y = 0

⇥ⓘ To add fractions, put them over the same denominator, then add the numerators:

\frac{6 x + 4}{x - 1} - y

\frac{(6 x + 4) - y \cdot (x - 1)}{x - 1}

⇥

\frac{(6 x + 4) - y \cdot (x - 1)}{x - 1} = 0

⇥

\frac{6 x + 4 - y \cdot (x - 1)}{x - 1} = 0

⇥A quotient is zero if and only if its numerator is zero and its denominator is not: we solve the numerator equal to 0

⇥⇥1.2.1) Let's solve in

ℝ

6 x + 4 - y \cdot (x - 1) = 0

⇥⇥ⓘ Expand by multiplying each term inside the parentheses:

- y \cdot (x - 1)

- (y x - y)

⇥⇥

6 x + 4 - (y x - y) = 0

⇥⇥ⓘ A minus sign before a parenthesis changes the sign of each of its terms:

- (y x - y)

- y \cdot x + y

⇥⇥

6 x

+ 4

- y \cdot x

+ y = 0

⇥⇥

6 x - y x = - 4 - y

⇥⇥

(6 - y) \cdot x = - 4 - y

⇥⇥ⓘ Both sides are divided by

(6 - y)

, which must be nonzero:

y \neq 6

⇥⇥

x = \frac{- 4 - y}{6 - y}

⇥⇥ⓘ The minus sign is factored out, changing the sign of each term:

- 4 - y

- (4 + y)

⇥⇥

x = - \frac{4 + y}{6 - y}

⇥⇥Solution found:

x = - \frac{y + 4}{6 - y}

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

x - 1 \neq 0

⇥⇥Domain condition:

x \neq 1

Renaming the output as

x

, the inverse is

- \frac{x + 4}{6 - x}

, with domain

x \neq 6

, the condition found when isolating the input

▶Inverse function:

- \frac{x + 4}{6 - x}