The function \(f\) is defined by \(f : x \mapsto \frac{x+3}{2x-1}\), \(x \in \mathbb{R}, x \neq \frac{1}{2}\).
(i) Show that \(ff(x) = x\).
(ii) Hence, or otherwise, obtain an expression for \(f^{-1}(x)\).
Solution
(i) To find \(ff(x)\), we substitute \(f(x)\) into itself:
\(f(x) = \frac{x+3}{2x-1}\)
\(ff(x) = f\left(\frac{x+3}{2x-1}\right) = \frac{\frac{x+3}{2x-1} + 3}{2\left(\frac{x+3}{2x-1}\right) - 1}\)
Simplifying the numerator:
\(\frac{x+3}{2x-1} + 3 = \frac{x+3 + 3(2x-1)}{2x-1} = \frac{7x}{2x-1}\)
Simplifying the denominator:
\(2\left(\frac{x+3}{2x-1}\right) - 1 = \frac{2(x+3)}{2x-1} - 1 = \frac{2x+6 - (2x-1)}{2x-1} = \frac{7}{2x-1}\)
Thus, \(ff(x) = \frac{7x}{7} = x\).
(ii) To find \(f^{-1}(x)\), we set \(y = \frac{x+3}{2x-1}\) and solve for \(x\):
\(y(2x-1) = x+3\)
\(2xy - y = x + 3\)
\(x(2y-1) = y + 3\)
\(x = \frac{y+3}{2y-1}\)
Thus, \(f^{-1}(x) = \frac{x+3}{2x-1}\).
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