A function is defined by \(h : x \to 2 - \frac{1}{2x - 3}, x \neq \frac{3}{2}\). Find \(h^{-1}(\frac{1}{2})\).

A.

\(6\)

B.

\(\frac{11}{6}\)

C.

\(\frac{11}{4}\)

D.

\(\frac{5}{3}\)

Correct answer is B

\(h : x \to 2 - \frac{1}{2x - 3}\)

\(h(x) = \frac{2(2x - 3) - 1}{2x - 3} = \frac{4x - 7}{2x - 3}\)

Let x = h(y)

\(x = \frac{4y - 7}{2y - 3}\)

\(x(2y - 3) = 4y - 7  \implies 2xy - 4y = 3x - 7\)

\(y = \frac{3x - 7}{2x - 4}\)

\(h^{-1}(x) = \frac{3x - 7}{2x - 4}\)

\(\therefore  h^{-1}(\frac{1}{2}) = \frac{3(\frac{1}{2}) - 7}{2(\frac{1}{2}) - 4}\)

= \(\frac{\frac{-11}{2}}{-3} = \frac{11}{6}\)