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

A.

\(\frac{3x - 4}{2x - 7}, x \neq \frac{7}{2}\)

B.

\(\frac{3x - 7}{2x - 4}, x \neq 2\)

C.

\(\frac{2x - 7}{4x - 3}, x \neq \frac{3}{4}\)

D.

\(\frac{4x - 7}{2x - 4}, x \neq 2\)

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)

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

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

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

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