To determine the restricted domain that makes the given function \( f(x) = (x - 5)^2 + 2 \) one-to-one, we first recognize that this is a parabolic function that opens upwards. The vertex of the parabola is at the point \( (5, 2) \).
To make \( f(x) \) one-to-one, we can restrict the domain to either \( x \geq 5 \) or \( x \leq 5 \). However, the typical choice is to go with \( x \geq 5 \) so that the function is increasing.
Restricted Domain:
\[ x \geq 5 \]
Finding the Inverse Function:
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Set \( y = f(x) \): \[ y = (x - 5)^2 + 2 \]
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Solve for \( x \): \[ y - 2 = (x - 5)^2 \]
\[ \sqrt{y - 2} = x - 5 \] (taking the positive square root, since we're restricting \( x \) to values \( \geq 5 \))
\[ x = \sqrt{y - 2} + 5 \]
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Interchanging \( x \) and \( y \) to find the inverse: \[ f^{-1}(x) = \sqrt{x - 2} + 5 \]
Thus, the complete solution is:
- Restricted domain: \( x \geq 5 \)
- Inverse function: \( f^{-1}(x) = \sqrt{x - 2} + 5 \)
However, note that the responses given do not include this exact form and are differences on the format of the inverse function. Let's check your options in the responses:
- Restricted domain: \( x \geq 5; f^{-1}(x) = 5 - \sqrt{x - 2} \)
- Restricted domain: \( x \geq 5; f^{-1}(x) = 5 + \sqrt{x - 2} \) (this is correct)
- Restricted domain: \( x \leq 5; f^{-1}(x) = 5 + \sqrt{x + 2} \)
- Restricted domain: \( x \leq -5; f^{-1}(x) = 5 + \sqrt{x - 2} \)
The correct response based on our calculations is: restricted domain: \( x \geq 5; f^{-1}(x) = 5 + \sqrt{x - 2} \)** (Option 2).
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