Find the inverse function of f(x) = (2x+3)/(x-1)

Question

Find the inverse function of $f(x) = \frac{2x+3}{x-1}$ . Also find the domain of $f^{-1}$ .

Solution — Step by Step

Let $y = f(x) = \frac{2x+3}{x-1}$

We want to express $x$ in terms of $y$ :

y(x - 1) = 2x + 3

xy - y = 2x + 3

xy - 2x = 3 + y

x(y - 2) = y + 3

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

Replacing $y$ with $x$ (swapping the variable name to standard form):

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

The domain of $f^{-1}(x) = \frac{x+3}{x-2}$ is all real numbers except where the denominator is zero:

$x - 2 = 0 \Rightarrow x = 2$

Domain of $f^{-1}$ : $\mathbb{R} \setminus \{2\}$ (all reals except 2)

Note: The domain of $f^{-1}$ = the range of $f$ . Let’s verify: $f(x) = \frac{2x+3}{x-1}$ approaches 2 as $x \to \infty$ (since ratio of leading coefficients = 2/1 = 2), and can never equal 2 exactly (since $\frac{2x+3}{x-1} = 2 \Rightarrow 2x+3 = 2x-2 \Rightarrow 3 = -2$ , impossible). So range of $f$ = $\mathbb{R} \setminus \{2\}$ ✓

$(f \circ f^{-1})(x) = f\left(\frac{x+3}{x-2}\right)$

= \frac{2\left(\frac{x+3}{x-2}\right) + 3}{\left(\frac{x+3}{x-2}\right) - 1}

Numerator: $\frac{2(x+3)}{x-2} + 3 = \frac{2x+6 + 3(x-2)}{x-2} = \frac{2x+6+3x-6}{x-2} = \frac{5x}{x-2}$

Denominator: $\frac{x+3}{x-2} - 1 = \frac{x+3-(x-2)}{x-2} = \frac{5}{x-2}$

f(f^{-1}(x)) = \frac{5x/(x-2)}{5/(x-2)} = \frac{5x}{5} = x \checkmark

Why This Works

Finding an inverse means “undoing” the function. If $f$ maps $a$ to $b$ , then $f^{-1}$ maps $b$ back to $a$ . The procedure — set $y = f(x)$ , solve for $x$ , then replace $y$ with $x$ — is the standard algebraic method. The verification step ( $f \circ f^{-1} =$ identity) confirms we haven’t made an error.

This particular function is a Möbius transformation (rational function of the form $(ax+b)/(cx+d)$ ). Möbius transformations always have themselves as their own inverse (when the transformation is an involution). Notice: $f^{-1}(x) = \frac{x+3}{x-2}$ has the same structure as $f(x) = \frac{2x+3}{x-1}$ — both are linear-over-linear.

A quick way to find the inverse of $f(x) = \frac{ax+b}{cx+d}$ : swap $a$ and $d$ , negate $b$ and $c$ : $f^{-1}(x) = \frac{dx-b}{-cx+a}$ . Verify: $f^{-1}(x) = \frac{1 \cdot x + 3}{-(-1)x + 2} = \frac{x+3}{x+2}$ … wait, let me apply properly. For $f(x) = \frac{2x+3}{x-1}$ : $a=2, b=3, c=1, d=-1$ . $f^{-1}(x) = \frac{-x-3}{-x+2} = \frac{x+3}{x-2}$ (multiply numerator and denominator by $-1$ ). ✓ This shortcut saves time in JEE.

Common Mistake

The most common error is writing $f^{-1}(x) = \frac{1}{f(x)} = \frac{x-1}{2x+3}$ . This is the reciprocal of $f$ , not the inverse. These are completely different things. $f^{-1}$ means “the function that undoes $f$ ”; it is NOT $1/f(x)$ . The composition $f(f^{-1}(x)) = x$ is the defining property of inverse — always use this to check your answer.

Question

Solution — Step by Step

Why This Works

Common Mistake

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