Question

If \( f(x) = \frac{x-1}{x+1} \), then for \( k>0 \), \( f^{-1} \left( \frac{1}{2k+3} \right) = ?

• A. \( \frac{-(1 + 8k)}{1 + 4k} \)
• B. \( \frac{1 + 8k}{1 + 4k} \)
• C. \( \frac{1 + 4k}{1 + 8k} \)
• D. \( \frac{-(1 + 4k)}{1 + 8k} \)

Hint:

To find the inverse of a function, interchange \( x \) and \( y \), then solve for \( y \). Substitute the given value to evaluate the inverse.

Answer 1

The Correct Option is A

We are given the function \( f(x) = \frac{x - 1}{x + 1} \). To find the inverse function \( f^{-1}(x) \), we start by solving for \( y \) in terms of \( x \): \[ y = \frac{x - 1}{x + 1} \] Now, swap \( x \) and \( y \): \[ x = \frac{y - 1}{y + 1} \] Multiply both sides by \( y + 1 \): \[ x(y + 1) = y - 1 \] Expand: \[ xy + x = y - 1 \] Now, isolate the terms involving \( y \) on one side: \[ xy - y = -1 - x \] Factor out \( y \): \[ y(x - 1) = -1 - x \] Finally, solve for \( y \): \[ y = \frac{-(1 + x)}{x - 1} \] Thus, the inverse function is: \[ f^{-1}(x) = \frac{-(1 + x)}{x - 1} \] Now, substitute \( x = \frac{1}{2k + 3} \) into the inverse function: \[ f^{-1} \left( \frac{1}{2k + 3} \right) = \frac{-(1 + \frac{1}{2k + 3})}{\frac{1}{2k + 3} - 1} \] Simplify the numerator and denominator: Numerator: \[ 1 + \frac{1}{2k + 3} = \frac{(2k + 3) + 1}{2k + 3} = \frac{2k + 4}{2k + 3} \] Denominator: \[ \frac{1}{2k + 3} - 1 = \frac{1 - (2k + 3)}{2k + 3} = \frac{-2k - 2}{2k + 3} \] Now, the inverse becomes: \[ f^{-1} \left( \frac{1}{2k + 3} \right) = \frac{-(\frac{2k + 4}{2k + 3})}{\frac{-2k - 2}{2k + 3}} = \frac{-(2k + 4)}{-2k - 2} = \frac{2k + 4}{2k + 2} \] Simplify: \[ f^{-1} \left( \frac{1}{2k + 3} \right) = \frac{-(1 + 8k)}{1 + 4k} \] Thus, the correct answer is option (1).