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If \(f(x)=\frac{1-x}{1+x}\) the domain of \(f^{-1}(x)\) is
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Verified Answer
The correct answer is:
\(\mathrm{R}-\{-1\}\)
Let \(f(x)=y\). Then \(\frac{1-x}{1+x}=y\)
\(\Rightarrow x=\frac{1-y}{1+y} \Rightarrow f^{-1}(y)=\frac{1-y}{1+y}\)
Thus, \(f^{-1}(x)=\frac{1-x}{1+x}\)
Clearly, \(\mathrm{f}^{-1}(\mathrm{x})\) is defined for \(1+\mathrm{x} \neq 0\).
Hence, domain of \(\mathrm{f}^{-1}(\mathrm{x})\) is \(\mathrm{R}-\{-1\}\)
\(\Rightarrow x=\frac{1-y}{1+y} \Rightarrow f^{-1}(y)=\frac{1-y}{1+y}\)
Thus, \(f^{-1}(x)=\frac{1-x}{1+x}\)
Clearly, \(\mathrm{f}^{-1}(\mathrm{x})\) is defined for \(1+\mathrm{x} \neq 0\).
Hence, domain of \(\mathrm{f}^{-1}(\mathrm{x})\) is \(\mathrm{R}-\{-1\}\)
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