Find the inverse function of f(x)=(2x-7)/(5+x) and state its domain and range.

f^-1(x)=(7+5x)/(2 - x), domain x≠ 2, range y≠ -5

Find the inverse function of $f (x) = \frac{2 x - 7}{5 + x}$ and state its domain and range.

hard 4 marks Class 12 · Mathematics · Relations and Functions

Answer

f^{- 1} (x) = \frac{7 + 5 x}{2 - x}

, domain

x \neq = 2

, range

y \neq = - 5

The correct answer is $f^{- 1} (x) = \frac{7 + 5 x}{2 - x}$ , domain $x \neq = 2$ , range $y \neq = - 5$ — Since the domain of $f$ is $R ∖ {- 5}$ , the range of $f^{- 1}$ (which equals the domain of $f$ ) is $R ∖ {- 5}$ .

Step-by-step solution

Start with $y = \frac{2 x - 7}{5 + x}$ and interchange $x$ and $y$ to solve for the inverse: $x = \frac{2 y - 7}{5 + y}$ .
Multiply both sides by the denominator: $x (5 + y) = 2 y - 7$ .
Expand: $5 x + x y = 2 y - 7$ .
Collect the terms containing $y$ on one side: $x y - 2 y = - 7 - 5 x$ .
Factor $y$ : $y (x - 2) = - (7 + 5 x)$ .
Solve for $y$ : $y = - \frac{7 + 5 x}{x - 2} = \frac{7 + 5 x}{2 - x}$ .
Thus the inverse function is $f^{- 1} (x) = \frac{7 + 5 x}{2 - x}$ .
The original function $f$ is undefined at $x = - 5$ (denominator $5 + x = 0$ ), so the domain of $f$ is $R ∖ {- 5}$ .
For $f^{- 1}$ , the denominator $2 - x$ must be non‑zero, giving $x \neq = 2$ ; hence the domain of $f^{- 1}$ (the range of $f$ ) is $R ∖ {2}$ .
Since the domain of $f$ is $R ∖ {- 5}$ , the range of $f^{- 1}$ (which equals the domain of $f$ ) is $R ∖ {- 5}$ .

Common mistake: Skipping the swap of

x

and

y

before solving, leading to incorrect algebra; also forgetting to exclude values that make denominators zero.

Concept tested: Inverse of a rational function with domain‑range analysis

Published 9 September 2026 · How this solution was prepared and checked

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Find the inverse function of f(x)=5+x2x−7​ and state its domain and range.

Step-by-step solution

Find the inverse function of $f (x) = \frac{2 x - 7}{5 + x}$ and state its domain and range.