[tex]f(x) = \frac{4x + 1}{3} \\ {f}^{ - 1} (y) = x - - - (1) \\ y = \frac{4x + 1}{3} \\ 3y = 4x + 1 \\ 4x = 3y - 1 \\ x = \frac{3y - 1}{4} putting \: value \: of \: x \\ in \: equation \: 1 \\ {f}^{ - 1} (y) = \frac{3y - 1}{4} \\ replace \: y \: by \: x \\ {f}^{ - 1} (x) = \frac{3x - 1}{4} - - - (2) \\ equation \: 2 \: is \: answer
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question 2
[tex]f(x) = \frac{x + 2}{x - 1} \\ {f}^{ - 1}(y) = x - - - - - (1) \\ y = \frac{x + 2}{x - 1} \\ y(x - 1) = x + 2 \\ yx - y = x + 2 \\ yx - x = y + 2 \\ x(y - 1) = y + 2 \\ \frac{x(y - 1)}{y - 1} = \frac{y + 2}{y - 1} \\ x = \frac{y + 2}{y - 1} \\ putting \: the \: value \: of \: x \: in \: 1 \\ {f}^{ - 1}(y) = \frac{y + 2}{y - 1} - - -( 2) \\ replace \: y \: by \: x \: in \: equation \: 2 \\ {f}^{ - 1}(x) = \frac{x + 2}{x - 1} [/tex]
