Question:

Let \( f(x) = \frac{3x - 5}{2x + 1} \). If \( f^{-1}(x) = \frac{x + a}{bx + c} \), then what is the value of \( (a - b + c) \)?

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When finding the inverse of a function, remember to swap the roles of \( x \) and \( y \), and then solve for the new \( y \).
Updated On: Apr 17, 2025
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The Correct Option is D

Solution and Explanation

We are given that \( f(x) = \frac{3x - 5}{2x + 1} \), and we need to find the inverse of this function.
To find \( f^{-1}(x) \), first, replace \( f(x) \) with \( y \): \[ y = \frac{3x - 5}{2x + 1} \] Now, solve for \( x \) in terms of \( y \): \[ y(2x + 1) = 3x - 5 \] \[ 2xy + y = 3x - 5 \] \[ 2xy - 3x = -y - 5 \] \[ x(2y - 3) = -y - 5 \] \[ x = \frac{-y - 5}{2y - 3} \] Now, replace \( y \) with \( x \) to get the inverse function: \[ f^{-1}(x) = \frac{x + a}{bx + c} \] By comparing the expressions, we find: \[ a = -5, \quad b = 2, \quad c = -3 \] Thus, \( a - b + c = -5 - 2 - 3 = -10 \).
So, the value of \( (a - b + c) \) is 10.
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