inverse of f(x)=(4x)/(3-7x) — 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,

4 \cdot 3 - 0 \cdot (-7)

12

≠ 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}{3 - 7 x}

⇥1.1) Let's solve in

ℝ

\frac{4 x}{3 - 7 x} = y

⇥1.2) Let's add fractions:

\frac{4 x}{3 - 7 x} - y = 0

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

\frac{4 x}{3 - 7 x} - y

\frac{4 x - y \cdot (3 - 7 x)}{3 - 7 x}

⇥

\frac{4 x - y \cdot (3 - 7 x)}{3 - 7 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

4 x - y \cdot (3 - 7 x) = 0

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

- y \cdot (3 - 7 x)

- (y \cdot 3 + y \cdot (-7) x)

⇥⇥

4 x - (y \cdot 3 + y \cdot (-7) x) = 0

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

- (y \cdot 3 + y \cdot (-7) x)

- y \cdot 3 - y \cdot (-7) x

⇥⇥

4 x - y \cdot 3 - y \cdot (-7) x = 0

⇥⇥ⓘ Two minus signs cancel out:

- y \cdot (-7)

7 y

⇥⇥

4 x - y \cdot 3 + 7 y x = 0

⇥⇥

4 x + 7 y x = y \cdot 3

⇥⇥

(4 + 7 y) \cdot x = y \cdot 3

⇥⇥ⓘ Both sides are divided by

(4 + 7 y)

, which must be nonzero:

y \neq - \frac{4}{7}

⇥⇥Solution found:

x = \frac{3 y}{7 y + 4}

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

3 - 7 x \neq 0

⇥⇥

7 x \neq 3

⇥⇥ⓘ Both sides are divided by

7

⇥⇥Domain condition:

x \neq \frac{3}{7}

Renaming the output as

x

, the inverse is

\frac{3 x}{7 x + 4}

, with domain

x \neq - \frac{4}{7}

, the condition found when isolating the input

▶Inverse function:

\frac{3 x}{7 x + 4}