inverse of g(x)=(-x-2)/(x+4) — step-by-step solution

1) Let's sum the fractions:

\frac{- x - 2}{x + 4}

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

- x - 2

- (x + 2)

- \frac{x + 2}{x + 4}

2) 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,

4 - 2

2

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

Express input

x

in terms of output

y

y = - \frac{x + 2}{x + 4}

⇥2.1) Let's solve in

ℝ

- \frac{x + 2}{x + 4} = y

⇥2.2) Let's add fractions:

- \frac{x + 2}{x + 4} - y = 0

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

- \frac{x + 2}{x + 4} - y

\frac{- (x + 2) - y \cdot (x + 4)}{x + 4}

⇥

\frac{- (x + 2) - y \cdot (x + 4)}{x + 4} = 0

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

- (x + 2)

- x - 2

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

- x - 2 - y \cdot (x + 4)

- (x + 2 + y \cdot (x + 4))

⇥

- \frac{x + 2 + 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

⇥⇥2.2.1) Let's solve in

ℝ

- (x + 2 + y \cdot (x + 4)) = 0

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

- (x + 2 + y \cdot (x + 4))

- x - 2 - y \cdot (x + 4)

⇥⇥

- x - 2 - y \cdot (x + 4)

⇥⇥

- x - 2 - y \cdot (x + 4) = 0

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

- y \cdot (x + 4)

- (y x + y \cdot 4)

⇥⇥

- x - 2 - (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

⇥⇥

- x

- 2

- y \cdot x

- y \cdot 4 = 0

⇥⇥ⓘ We add

2 + y \cdot 4

to both sides

⇥⇥

- x - y x = 2 + y \cdot 4

⇥⇥

(- 1 - y) \cdot x = 2 + y \cdot 4

⇥⇥ⓘ Both sides are divided by

(- 1 - y)

, which must be nonzero:

y \neq -1

⇥⇥

x = \frac{2 + y \cdot 4}{- 1 - y}

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

- 1 - y

- (1 + y)

⇥⇥

x = \frac{2 + y \cdot 4}{- (1 + y)}

⇥⇥ⓘ The sign of a fraction goes in front of the bar, whether it comes from the numerator or the denominator:

\frac{2 + y \cdot 4}{- (1 + y)}

- \frac{2 + y \cdot 4}{1 + y}

⇥⇥

x = - \frac{2 + y \cdot 4}{1 + y}

⇥⇥Solution obtained:

x = - \frac{4 y + 2}{y + 1}

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

x + 4 \neq 0

⇥⇥Domain condition:

x \neq -4

Renaming the output as

x

, the inverse is

- \frac{4 x + 2}{x + 1}

, with domain

x \neq -1

, the condition found when isolating the input

▶Inverse function:

- \frac{4 x + 2}{x + 1}