To find the inverse of the function \( f(x) = 2x + 1 \), we follow these steps:
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Replace \( f(x) \) with \( y \): \[ y = 2x + 1 \]
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Swap \( x \) and \( y \): \[ x = 2y + 1 \]
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Solve for \( y \): \[ x - 1 = 2y \ y = \frac{x - 1}{2} \]
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The inverse function \( f^{-1}(x) \) is: \[ f^{-1}(x) = \frac{x - 1}{2} \]
This can be rewritten in a more standard form: \[ f^{-1}(x) = \frac{1}{2}x - \frac{1}{2} \]
Now, let's compare this with the given options for \( h(x) \):
- \( h(x) = \frac{1}{2}x - \frac{1}{2} \)
- \( h(x) = \frac{1}{2}x + \frac{1}{2} \)
- \( h(x) = \frac{1}{2}x - 2 \)
- \( h(x) = \frac{1}{2}x + 2 \)
The correct inverse is: \[ h(x) = \frac{1}{2}x - \frac{1}{2} \]