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If $\mathrm{f}(x)=\frac{2 x-3}{3 x-4}, x \neq \frac{4}{3}$, then the value of $\mathrm{f}^{-1}(x)$ is
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Verified Answer
The correct answer is:
$\frac{4 x-3}{3 x-2}$
Let $\mathrm{f}(x)=y \Rightarrow x=\mathrm{f}^{-1}(y)$
$\begin{aligned}
& y=\frac{2 x-3}{3 x-4} \\
& \Rightarrow 3 x y-4 y=2 x-3 \\
& \Rightarrow x(3 y-2)=4 y-3 \\
& \Rightarrow x=\frac{4 y-3}{3 y-2} \\
& \Rightarrow \mathrm{f}^{-1}(y)=\frac{4 y-3}{3 y-2} \\
& \Rightarrow \mathrm{f}^{-1}(x)=\frac{4 x-3}{3 x-2}
\end{aligned}$
$\begin{aligned}
& y=\frac{2 x-3}{3 x-4} \\
& \Rightarrow 3 x y-4 y=2 x-3 \\
& \Rightarrow x(3 y-2)=4 y-3 \\
& \Rightarrow x=\frac{4 y-3}{3 y-2} \\
& \Rightarrow \mathrm{f}^{-1}(y)=\frac{4 y-3}{3 y-2} \\
& \Rightarrow \mathrm{f}^{-1}(x)=\frac{4 x-3}{3 x-2}
\end{aligned}$
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