A function is defined by \(h : x \to 2 - \frac{1}{2x - 3}, x \neq \frac{3}{2}\). Find \(h^{-1}(\frac{1}{2})\).
\(h : x \to 2 - \frac{1}{2x - 3}\)
\(h(x) = \frac{2(2x - 3) - 1}{2x - 3} = \frac{4x - 7}{2x - 3}\)
Let x = h(y)
\(x = \frac{4y - 7}{2y - 3}\)
\(x(2y - 3) = 4y - 7 \implies 2xy - 4y = 3x - 7\)
\(y = \frac{3x - 7}{2x - 4}\)
\(h^{-1}(x) = \frac{3x - 7}{2x - 4}\)
\(\therefore h^{-1}(\frac{1}{2}) = \frac{3(\frac{1}{2}) - 7}{2(\frac{1}{2}) - 4}\)
= \(\frac{\frac{-11}{2}}{-3} = \frac{11}{6}\)
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