Answer :
To address the problem of finding an inverse function for [tex]\( f(x) = -x^2 + 10 \)[/tex], we should approach it step-by-step. Here's a detailed solution:
### 1. Understanding the Problem:
The function [tex]\( f(x) = -x^2 + 10 \)[/tex] is not one-to-one over the entire real line since it is a downward-facing parabola. To find an inverse function, we first need to restrict the domain to a portion where [tex]\( f(x) \)[/tex] is one-to-one.
### 2. Identifying the Vertex:
The vertex of the parabola [tex]\( f(x) = -x^2 + 10 \)[/tex] serves as a key reference point. The vertex occurs at [tex]\( x = 0 \)[/tex], making the vertex [tex]\((0, 10)\)[/tex]. A function is one-to-one in an interval on either side of the vertex.
### 3. Choosing the Restricted Domain:
- For the function to be one-to-one, we can restrict the domain to the interval [tex]\([0, \infty)\)[/tex] because on this interval, [tex]\( f(x) \)[/tex] is monotonically decreasing and thus one-to-one.
### 4. Solving for the Inverse:
To find the inverse function, we solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex] where [tex]\( y = f(x) \)[/tex]:
[tex]\[ y = -x^2 + 10 \][/tex]
Rearrange to solve for [tex]\( x \)[/tex]:
[tex]\[ x^2 = 10 - y \][/tex]
Taking the positive square root (because our restricted domain is [tex]\( x \geq 0 \)[/tex]):
[tex]\[ x = \sqrt{10 - y} \][/tex]
For the inverse function, we swap [tex]\( y \)[/tex] and [tex]\( x \)[/tex]:
[tex]\[ f^{-1}(x) = \sqrt{10 - x} \][/tex]
### 5. Conclusion:
The restricted domain where [tex]\( f \)[/tex] is one-to-one is [tex]\([0, \infty)\)[/tex], and the corresponding inverse function is:
[tex]\[ f^{-1}(x) = \sqrt{10 - x} \][/tex]
Therefore, the correct solution is:
- The restricted domain for [tex]\( f \)[/tex] is [tex]\([0, \infty)\)[/tex] and
- The inverse function is [tex]\( f^{-1}(x) = \sqrt{10 - x} \)[/tex].
Thus, the correct answer is:
The restricted domain for [tex]\( f \)[/tex] is [tex]\( [0, \infty) \)[/tex] and [tex]\( f^{-1}(x) = \sqrt{10 - x} \)[/tex].
### 1. Understanding the Problem:
The function [tex]\( f(x) = -x^2 + 10 \)[/tex] is not one-to-one over the entire real line since it is a downward-facing parabola. To find an inverse function, we first need to restrict the domain to a portion where [tex]\( f(x) \)[/tex] is one-to-one.
### 2. Identifying the Vertex:
The vertex of the parabola [tex]\( f(x) = -x^2 + 10 \)[/tex] serves as a key reference point. The vertex occurs at [tex]\( x = 0 \)[/tex], making the vertex [tex]\((0, 10)\)[/tex]. A function is one-to-one in an interval on either side of the vertex.
### 3. Choosing the Restricted Domain:
- For the function to be one-to-one, we can restrict the domain to the interval [tex]\([0, \infty)\)[/tex] because on this interval, [tex]\( f(x) \)[/tex] is monotonically decreasing and thus one-to-one.
### 4. Solving for the Inverse:
To find the inverse function, we solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex] where [tex]\( y = f(x) \)[/tex]:
[tex]\[ y = -x^2 + 10 \][/tex]
Rearrange to solve for [tex]\( x \)[/tex]:
[tex]\[ x^2 = 10 - y \][/tex]
Taking the positive square root (because our restricted domain is [tex]\( x \geq 0 \)[/tex]):
[tex]\[ x = \sqrt{10 - y} \][/tex]
For the inverse function, we swap [tex]\( y \)[/tex] and [tex]\( x \)[/tex]:
[tex]\[ f^{-1}(x) = \sqrt{10 - x} \][/tex]
### 5. Conclusion:
The restricted domain where [tex]\( f \)[/tex] is one-to-one is [tex]\([0, \infty)\)[/tex], and the corresponding inverse function is:
[tex]\[ f^{-1}(x) = \sqrt{10 - x} \][/tex]
Therefore, the correct solution is:
- The restricted domain for [tex]\( f \)[/tex] is [tex]\([0, \infty)\)[/tex] and
- The inverse function is [tex]\( f^{-1}(x) = \sqrt{10 - x} \)[/tex].
Thus, the correct answer is:
The restricted domain for [tex]\( f \)[/tex] is [tex]\( [0, \infty) \)[/tex] and [tex]\( f^{-1}(x) = \sqrt{10 - x} \)[/tex].
