What is the inverse of the function [tex]f(x) = 2x + 1[/tex]?

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



Answer :

To determine the inverse of the function \( f(x) = 2x + 1 \), we need to follow a systematic process. Let's go through the steps in detail:

1. Write the function in terms of \( y \):
[tex]\[ y = 2x + 1 \][/tex]

2. Swap \( x \) and \( y \):
To find the inverse, we swap \( x \) and \( y \):
[tex]\[ x = 2y + 1 \][/tex]

3. Solve for \( y \):
Now, we need to solve this equation for \( y \). Start by isolating \( y \):
[tex]\[ x - 1 = 2y \][/tex]
[tex]\[ y = \frac{x - 1}{2} \][/tex]

4. Simplify the expression:
The equation can be rewritten as:
[tex]\[ y = \frac{1}{2}x - \frac{1}{2} \][/tex]

5. Express the inverse function:
The inverse function \( f^{-1}(x) \) is then:
[tex]\[ f^{-1}(x) = \frac{1}{2}x - \frac{1}{2} \][/tex]

Now we compare this result with the given options:

1. \( h(x) = \frac{1}{2}x - \frac{1}{2} \)
2. \( h(x) = \frac{1}{2}x + \frac{1}{2} \)
3. \( h(x) = \frac{1}{2}x - 2 \)
4. \( h(x) = \frac{1}{2}x + 2 \)

Clearly, the first option \( h(x) = \frac{1}{2}x - \frac{1}{2} \) matches our derived inverse function.

Thus, the inverse function is:
[tex]\[ h(x) = \frac{1}{2}x - \frac{1}{2} \][/tex]

So, the correct answer is the first option:
[tex]\[ h(x) = \frac{1}{2}x - \frac{1}{2} \][/tex]