inverse of f(x)=((4x-1))/((2x+3))

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,

4 \cdot 3 - (-1 \cdot 2)

14

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

Express input

x

in terms of output

y

y = \frac{4 x - 1}{2 x + 3}

⇥1.1) Let's solve in

ℝ

\frac{4 x - 1}{2 x + 3} = y

⇥1.2) Let's sum the fractions:

\frac{4 x - 1}{2 x + 3} - y = 0

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

\frac{4 x - 1}{2 x + 3} - y

\frac{(4 x - 1) - y \cdot (2 x + 3)}{2 x + 3}

⇥

\frac{(4 x - 1) - y \cdot (2 x + 3)}{2 x + 3} = 0

⇥

\frac{4 x - 1 - y \cdot (2 x + 3)}{2 x + 3} = 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

4 x - 1 - y \cdot (2 x + 3) = 0

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

- y \cdot (2 x + 3)

- (y \cdot 2 x + y \cdot 3)

⇥⇥

4 x - 1 - (y \cdot 2 x + y \cdot 3) = 0

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

- (y \cdot 2 x + y \cdot 3)

- y \cdot 2 x - y \cdot 3

⇥⇥

4 x

- 1

- y \cdot 2 x

- y \cdot 3 = 0

⇥⇥

4 x - y \cdot 2 x = 1 + y \cdot 3

⇥⇥

(4 - y \cdot 2) \cdot x = 1 + y \cdot 3

⇥⇥ⓘ Both sides are divided by

(4 - y \cdot 2)

, which must be nonzero:

y \neq 2

⇥⇥Solution obtained:

x = \frac{3 y + 1}{4 - 2 y}

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

2 x + 3 \neq 0

⇥⇥

2 x \neq -3

⇥⇥ⓘ Both sides are divided by

2

⇥⇥Domain condition:

x \neq - \frac{3}{2}

Renaming the output as

x

, the inverse is

\frac{3 x + 1}{4 - 2 x}

, with domain

x \neq 2

, the condition found when isolating the input

▶Inverse function:

\frac{3 x + 1}{4 - 2 x}