If f: R - {frac{3}{5}} rightarrow R - {frac{3 | vedclass

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(B) To find the inverse of the function $f(x) = \frac{3x+1}{5x-3}$,let $y = f(x)$.So,$y = \frac{3x+1}{5x-3}$.Cross-multiplying gives $y(5x-3) = 3x+1$,

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$f^{-1}(x) = f(x)$

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(B) To find the inverse of the function $f(x) = \frac{3x+1}{5x-3}$,let $y = f(x)$.So,$y = \frac{3x+1}{5x-3}$.Cross-multiplying gives $y(5x-3) = 3x+1$,which simplifies to $5xy - 3y = 3x + 1$.Rearranging the terms to solve for $x$: $5xy - 3x = 3y + 1$.Factoring out $x$: $x(5y - 3) = 3y + 1$.Thus,$x = \frac{3y+1}{5y-3}$.By definition,$f^{-1}(y) = x = \frac{3y+1}{5y-3}$.Replacing $y$ with $x$,we get $f^{-1}(x) = \frac{3x+1}{5x-3}$.Since $f^{-1}(x) = f(x)$,the correct option is $B$.

If $f: R - \{\frac{3}{5}\} \rightarrow R - \{\frac{3}{5}\}$ is defined by $f(x) = \frac{3x+1}{5x-3}$,then which of the following is true?

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