If f(x) = frac{2x - 1}{x + 5} (x ne -5),then | vedclass

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(B) Let $f(x) = y$,which implies $x = f^{-1}(y)$.Given $y = \frac{2x - 1}{x + 5}$,where $x 
e -5$.Multiply both sides by $(x + 5)$:$y(x + 5) = 2x - 1

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$\frac{5x + 1}{2 - x}, \; x 
e 2$

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(B) Let $f(x) = y$,which implies $x = f^{-1}(y)$.Given $y = \frac{2x - 1}{x + 5}$,where $x 
e -5$.Multiply both sides by $(x + 5)$:$y(x + 5) = 2x - 1$$xy + 5y = 2x - 1$Rearrange the terms to isolate $x$:$5y + 1 = 2x - xy$$5y + 1 = x(2 - y)$Solving for $x$:$x = \frac{5y + 1}{2 - y}$Since $x = f^{-1}(y)$,we have:$f^{-1}(y) = \frac{5y + 1}{2 - y}$Replacing $y$ with $x$,we get:$f^{-1}(x) = \frac{5x + 1}{2 - x}$,where $x 
e 2$.

If $f(x) = \frac{2x - 1}{x + 5}$ $(x \ne -5)$,then $f^{-1}(x)$ is equal to

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