Find the inverse function of h(x)=(4x+1)/(2-x). State the domain and range of h^-1(x). (Give the inverse in the form h^-1(x)=....)

h^-1(x)=(2x-1)/(4+x), domain x≠ -4, range y≠ 2

Find the inverse function of $h (x) = \frac{4 x + 1}{2 - x}$ . State the domain and range of $h^{- 1} (x)$ . (Give the inverse in the form $h^{- 1} (x) = \dots$ .)

hard 4 marks Class 12 · Mathematics · Relations and Functions

Answer

h^{- 1} (x) = \frac{2 x - 1}{4 + x}, domain x \neq = - 4, range y \neq = 2

The correct answer is $h^{- 1} (x) = \frac{2 x - 1}{4 + x}, domain x \neq = - 4, range y \neq = 2$ — Since the range of $h^{- 1}$ equals the domain of $h$ , find the domain of $h$ : denominator $2 - x \neq = 0 \Rightarrow x \neq = 2$ ; hence the range of $h^{- 1}$ is $y \neq = 2$ .

Step-by-step solution

Write $y = h (x) = \frac{4 x + 1}{2 - x}$ and interchange $x$ and $y$ to solve for $y$ :
Set $x = \frac{4 y + 1}{2 - y}$ .
Cross‑multiply: $x (2 - y) = 4 y + 1$ .
Expand: $2 x - x y = 4 y + 1$ .
Collect terms containing $y$ on one side: $- x y - 4 y = 1 - 2 x$ .
Factor $y$ : $y (- x - 4) = 1 - 2 x$ .
Solve for $y$ : $y = \frac{1 - 2 x}{- x - 4} = \frac{2 x - 1}{x + 4}$ (multiply numerator and denominator by $- 1$ ).
Thus $h^{- 1} (x) = \frac{2 x - 1}{4 + x}$ .
Domain of $h^{- 1}$ consists of all real numbers except where its denominator $4 + x = 0$ , i.e., $x \neq = - 4$ .
Since the range of $h^{- 1}$ equals the domain of $h$ , find the domain of $h$ : denominator $2 - x \neq = 0 \Rightarrow x \neq = 2$ ; hence the range of $h^{- 1}$ is $y \neq = 2$ .

Common mistake: Algebraic sign error while isolating

y

, leading to an incorrect inverse.

Concept tested: Finding inverse and swapping domain‑range

Published 12 September 2026 · How this solution was prepared and checked

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Find the inverse function of h(x)=2−x4x+1​. State the domain and range of h−1(x). (Give the inverse in the form h−1(x)=….)

Step-by-step solution

Find the inverse function of $h (x) = \frac{4 x + 1}{2 - x}$ . State the domain and range of $h^{- 1} (x)$ . (Give the inverse in the form $h^{- 1} (x) = \dots$ .)