Find the inverse function of h(x)=(3x+2)/(,1-2x,) and state its domain and range.

h^-1(x)=(x-2)/(2x+3), domain Rsetminus1/2, range Rsetminus-3/2

Find the inverse function of $h (x) = \frac{3 x + 2}{, 1 - 2 x ,}$ and state its domain and range.

hard 5 marks Class 12 · Mathematics · Relations and Functions

Answer

h^{- 1} (x) = \frac{x - 2}{2 x + 3}

, domain

R se t min u s \frac{1}{2}

, range

R se t min u s - \frac{3}{2}

The correct answer is $h^{- 1} (x) = \frac{x - 2}{2 x + 3}$ , domain $R se t min u s \frac{1}{2}$ , range $R se t min u s - \frac{3}{2}$ — Therefore domain of $h^{- 1}$ is all real $x$ except $x = \frac{1}{2}$ and range of $h^{- 1}$ is all real except $- \frac{3}{2}$ .

Step-by-step solution

Let $y = h (x) = \frac{3 x + 2}{1 - 2 x}$ .
Multiply both sides by $1 - 2 x$ : $y (1 - 2 x) = 3 x + 2$ .
Expand: $y - 2 y x = 3 x + 2$ .
Collect $x$ terms: $- 2 y x - 3 x = 2 - y$ .
Factor $x$ : $x (- 2 y - 3) = 2 - y$ .
Solve for $x$ : $x = \frac{2 - y}{- 2 y - 3} = \frac{y - 2}{2 y + 3}$ .
Thus $h^{- 1} (y) = \frac{y - 2}{2 y + 3}$ . Replace $y$ by $x$ for the inverse: $h^{- 1} (x) = \frac{x - 2}{2 x + 3}$ .
Domain of $h$ : $x e q \frac{1}{2}$ (where denominator $1 - 2 x = 0$ ).
Range of $h$ : all real $y$ except $y = - \frac{3}{2}$ (horizontal asymptote; no $x$ satisfies $y = - \frac{3}{2}$ ).
Therefore domain of $h^{- 1}$ is all real $x$ except $x = \frac{1}{2}$ and range of $h^{- 1}$ is all real except $- \frac{3}{2}$ .

Common mistake: Mis‑solving the linear equation for

x

Concept tested: Finding inverse of rational function

Published 15 September 2026 · How this solution was prepared and checked

← All Relations and Functions questions

Find the inverse function of h(x)=,1−2x,3x+2​ and state its domain and range.

Step-by-step solution

Find the inverse function of $h (x) = \frac{3 x + 2}{, 1 - 2 x ,}$ and state its domain and range.