If f(x) = frac{x}{1 + x},then {f^{-1}}(x) is | vedclass

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(D) Given $f(x) = \frac{x}{1 + x}$.Let $y = f(x)$,then $x = {f^{-1}}(y)$.Substituting $y$ for $f(x)$,we have $y = \frac{x}{1 + x}$.Cross-multiplying g

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$\frac{x}{1 - x}$

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(D) Given $f(x) = \frac{x}{1 + x}$.Let $y = f(x)$,then $x = {f^{-1}}(y)$.Substituting $y$ for $f(x)$,we have $y = \frac{x}{1 + x}$.Cross-multiplying gives $y(1 + x) = x$,which simplifies to $y + yx = x$.Rearranging the terms to solve for $x$,we get $y = x - yx = x(1 - y)$.Thus,$x = \frac{y}{1 - y}$.Since $x = {f^{-1}}(y)$,we have ${f^{-1}}(y) = \frac{y}{1 - y}$.Replacing $y$ with $x$,we obtain ${f^{-1}}(x) = \frac{x}{1 - x}$.

If $f(x) = \frac{x}{1 + x}$,then ${f^{-1}}(x)$ is equal to

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