inverse of f(x)=6/(5-x) — 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 5 - 6 \cdot (-1)

6

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

Express input

x

in terms of output

y

y = \frac{6}{5 - x}

⇥1.1) Let's solve in

ℝ

\frac{6}{5 - x} = y

⇥1.2) Let's sum the fractions:

\frac{6}{5 - x} - y = 0

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

\frac{6}{5 - x} - y

\frac{6 - y \cdot (5 - x)}{5 - x}

⇥

\frac{6 - y \cdot (5 - x)}{5 - x} = 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 - y \cdot (5 - x) = 0

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

- y \cdot (5 - x)

- (y \cdot 5 + y \cdot (- x))

⇥⇥

6 - (y \cdot 5 + y \cdot (- x)) = 0

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

- (y \cdot 5 + y \cdot (- x))

- y \cdot 5 - y \cdot (- x)

⇥⇥

6 - y \cdot 5 - y \cdot (- x) = 0

⇥⇥ⓘ Two minus signs cancel out:

- y \cdot (- x)

y x

⇥⇥

6 - y \cdot 5 + y x = 0

⇥⇥

y x = - 6 + y \cdot 5

⇥⇥ⓘ Both sides are divided by

y

, which must be nonzero:

y \neq 0

⇥⇥Solution obtained:

x = \frac{5 y - 6}{y}

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

5 - x \neq 0

⇥⇥Domain condition:

x \neq 5

Renaming the output as

x

, the inverse is

\frac{5 x - 6}{x}

, with domain

x \neq 0

, the condition found when isolating the input

▶Inverse function:

\frac{5 x - 6}{x}