inverse of f(x)=(3x+5)/(x-4) — 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 \cdot (-4) - 5

-17

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

Express input

x

in terms of output

y

y = \frac{3 x + 5}{x - 4}

⇥1.1) Let's solve in

ℝ

\frac{3 x + 5}{x - 4} = y

⇥1.2) Let's add fractions:

\frac{3 x + 5}{x - 4} - y = 0

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

\frac{3 x + 5}{x - 4} - y

\frac{(3 x + 5) - y \cdot (x - 4)}{x - 4}

⇥

\frac{(3 x + 5) - y \cdot (x - 4)}{x - 4} = 0

⇥

\frac{3 x + 5 - y \cdot (x - 4)}{x - 4} = 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

3 x + 5 - y \cdot (x - 4) = 0

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

- y \cdot (x - 4)

- (y x + y \cdot (-4))

⇥⇥

3 x + 5 - (y x + y \cdot (-4)) = 0

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

- (y x + y \cdot (-4))

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

⇥⇥

3 x

+ 5

- y \cdot x

- y \cdot (-4) = 0

⇥⇥ⓘ Two minus signs cancel out:

- y \cdot (-4)

4 y

⇥⇥

3 x

+ 5

- y x

+ 4 y = 0

⇥⇥

3 x - y x = - 5 - 4 y

⇥⇥

(3 - y) \cdot x = - 5 - 4 y

⇥⇥ⓘ Both sides are divided by

(3 - y)

, which must be nonzero:

y \neq 3

⇥⇥

x = \frac{- 5 - 4 y}{3 - y}

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

- 5 - 4 y

- (5 + 4 y)

⇥⇥

x = - \frac{5 + 4 y}{3 - y}

⇥⇥Solution found:

x = - \frac{4 y + 5}{3 - y}

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

x - 4 \neq 0

⇥⇥Domain condition:

x \neq 4

Renaming the output as

x

, the inverse is

- \frac{4 x + 5}{3 - x}

, with domain

x \neq 3

, the condition found when isolating the input

▶Inverse function:

- \frac{4 x + 5}{3 - x}