Question

The function f left parenthesis x right parenthesis equals left parenthesis x minus 5 right parenthesis squared plus 2 is not one-to-one. Identify a restricted domain that makes the function one-to-one, and find the inverse function.
(1 point)
Responses

restricted domain: x greater or equals 5; f superscript negative 1 baseline left parenthesis x right parenthesis equals 5 minus start root x minus 2 end root
restricted domain: Image with alt text: x greater or equals 5 ; Image with alt text: f superscript negative 1 baseline left parenthesis x right parenthesis equals 5 minus start root x minus 2 end root

restricted domain: x greater or equals 5; f superscript negative 1 baseline left parenthesis x right parenthesis equals 5 plus start root x minus 2 end root
restricted domain: Image with alt text: x greater or equals 5 ; Image with alt text: f superscript negative 1 baseline left parenthesis x right parenthesis equals 5 plus start root x minus 2 end root

restricted domain: x less than or equals 5; f superscript negative 1 baseline left parenthesis x right parenthesis equals 5 plus start root x plus 2 end root
restricted domain: Image with alt text: x less than or equals 5 ; Image with alt text: f superscript negative 1 baseline left parenthesis x right parenthesis equals 5 plus start root x plus 2 end root

restricted domain: x less than or equals negative 5; f superscript negative 1 baseline left parenthesis x right parenthesis equals 5 plus start root x minus 2 end root

Answer 1

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:
1. Set \( y = f(x) \):
\[ y = (x - 5)^2 + 2 \]

2. 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 \( \geq 5 \))

\[ x = \sqrt{y - 2} + 5 \]

3. 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:

1. **Restricted domain:** \( x \geq 5; f^{-1}(x) = 5 - \sqrt{x - 2} \)
2. **Restricted domain:** \( x \geq 5; f^{-1}(x) = 5 + \sqrt{x - 2} \) (this is correct)
3. **Restricted domain:** \( x \leq 5; f^{-1}(x) = 5 + \sqrt{x + 2} \)
4. **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).