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.
- \(\frac{3x - 4}{2x - 7}, x \neq \frac{7}{2}\)
-
\(\frac{3x - 7}{2x - 4}, x \neq 2\)
- \(\frac{2x - 7}{4x - 3}, x \neq \frac{3}{4}\)
- \(\frac{4x - 7}{2x - 4}, x \neq 2\)
The correct answer is: B
Explanation
\(h : x \to 2 - \frac{1}{2x - 3}\)
\(h(x) = \frac{2(2x - 3) - 1}{2x - 3} = \frac{4x - 7}{2x - 3}\)
Let x = h(y)
\(x = \frac{4y - 7}{2y - 3}\)
\(x(2y - 3) = 4y - 7 \implies 2xy - 4y = 3x - 7\)
\(y = \frac{3x - 7}{2x - 4}\)
\(h^{-1}(x) = \frac{3x - 7}{2x - 4}\)