Let [tex] f(x) = \frac{x+1}{x-2} [/tex] be a function defined on the set of real numbers excluding 2. Find the inverse of [tex] f [/tex].

Select one:

A. [tex] f^{-1}(x) = \frac{2x+1}{x-1} [/tex]
B. [tex] f^{-1}(x) = \frac{2x-1}{x-1} [/tex]
C. [tex] f^{-1}(x) = \frac{2x-1}{x+1} [/tex]
D. [tex] f^{-1}(x) = \frac{2x+1}{x+1} [/tex]



Answer :

To find the inverse of the function [tex]\( f(x) = \frac{x + 1}{x - 2} \)[/tex], we follow a series of steps. Let's denote the inverse function by [tex]\( f^{-1}(x) \)[/tex].

1. Express the function in terms of [tex]\( y \)[/tex] and [tex]\( x \)[/tex]:
[tex]\[ y = \frac{x + 1}{x - 2} \][/tex]

2. Swap [tex]\( y \)[/tex] and [tex]\( x \)[/tex] to solve for [tex]\( y \)[/tex]:
[tex]\[ x = \frac{y + 1}{y - 2} \][/tex]

3. Solve for [tex]\( y \)[/tex]:
First, multiply both sides by [tex]\( y - 2 \)[/tex] to clear the fraction:
[tex]\[ x(y - 2) = y + 1 \][/tex]

Distribute the [tex]\( x \)[/tex]:
[tex]\[ xy - 2x = y + 1 \][/tex]

Move all terms involving [tex]\( y \)[/tex] to one side and the constant terms to the opposite side:
[tex]\[ xy - y = 2x + 1 \][/tex]

Factor out [tex]\( y \)[/tex] on the left side:
[tex]\[ y(x - 1) = 2x + 1 \][/tex]

4. Isolate [tex]\( y \)[/tex]:
[tex]\[ y = \frac{2x + 1}{x - 1} \][/tex]

Therefore, the inverse function is:
[tex]\[ f^{-1}(x) = \frac{2x + 1}{x - 1} \][/tex]

5. Compare with the provided options:
The correct answer is:
[tex]\[ f^{-1}(x) = \frac{2x + 1}{x - 1} \][/tex]

Hence, the correct choice is:

[tex]\[ \boxed{ f^{-1}(x)=\frac{2x+1}{x-1} } \][/tex]

This corresponds to the first option provided.