To find the inverse of the function [tex]\( f(x) = 2x + 1 \)[/tex], we need to follow a series of steps.
1. Replace [tex]\( f(x) \)[/tex] with [tex]\( y \)[/tex]:
The original function is [tex]\( y = 2x + 1 \)[/tex].
2. Interchange [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
To find the inverse, swap the roles of [tex]\( x \)[/tex] and [tex]\( y \)[/tex]. So we write [tex]\( x = 2y + 1 \)[/tex].
3. Solve for [tex]\( y \)[/tex]:
To make [tex]\( y \)[/tex] the subject of the equation:
- Subtract 1 from both sides: [tex]\( x - 1 = 2y \)[/tex]
- Divide both sides by 2: [tex]\( y = \frac{x - 1}{2} \)[/tex]
4. Rewrite the inverse function:
Substitute [tex]\( y \)[/tex] with [tex]\( f^{-1}(x) \)[/tex], giving us the inverse function [tex]\( f^{-1}(x) = \frac{x - 1}{2} \)[/tex].
Simplifying the right-hand side, we get:
[tex]\[ f^{-1}(x) = \frac{1}{2} x - \frac{1}{2} \][/tex]
Now, let's compare this result with the given options:
1. [tex]\( h(x) = \frac{1}{2} x - \frac{1}{2} \)[/tex]
2. [tex]\( h(x) = \frac{1}{2} x + \frac{1}{2} \)[/tex]
3. [tex]\( h(x) = \frac{1}{2} x - 2 \)[/tex]
4. [tex]\( h(x) = \frac{1}{2} x + 2 \)[/tex]
The correct option that represents the inverse function [tex]\( f^{-1}(x) \)[/tex] is:
[tex]\[ h(x) = \frac{1}{2} x - \frac{1}{2} \][/tex]
Therefore, the inverse of the function [tex]\( f(x) = 2x + 1 \)[/tex] is [tex]\( h(x) = \frac{1}{2} x - \frac{1}{2} \)[/tex], which corresponds to the first option.
