To determine a portion of the domain where the function [tex]\( f(x) = x^2 - a \)[/tex] is one-to-one and to find the inverse for that restricted domain, let's go through the following steps:
1. Understanding the One-to-One Property:
- A function [tex]\( f(x) \)[/tex] is one-to-one if and only if each value of [tex]\( f(x) \)[/tex] corresponds to exactly one value of [tex]\( x \)[/tex]. In other words, [tex]\( f(x_1) = f(x_2) \)[/tex] implies [tex]\( x_1 = x_2 \)[/tex].
2. Analyzing the Original Function:
- The given function is [tex]\( f(x) = x^2 - a \)[/tex].
- Since this is a quadratic function, it is not one-to-one over its entire domain because quadratic functions are symmetric about their vertex, meaning they fail the horizontal line test.
3. Restricting the Domain:
- To make [tex]\( f(x) = x^2 - a \)[/tex] one-to-one, we can consider restricting [tex]\( x \)[/tex] to a domain where the function is either increasing or decreasing.
- For [tex]\( x^2 \)[/tex] type functions, [tex]\( x \)[/tex] restricted to [tex]\([0, \infty)\)[/tex] ensures that the function is monotonically increasing (positive values only).
4. Finding the Inverse Function:
- To find the inverse, we start with [tex]\( y = f(x) = x^2 - a \)[/tex].
- Solving for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]:
[tex]\[
y = x^2 - a
\][/tex]
[tex]\[
y + a = x^2
\][/tex]
[tex]\[
x = \sqrt{y + a}
\][/tex]
- Given the restricted domain [tex]\([0, \infty)\)[/tex], we keep the positive square root to satisfy the range of [tex]\( f \)[/tex] over this domain.
5. Conclusion of the Steps:
- Thus, for [tex]\( f(x) = x^2 - a \)[/tex] to be one-to-one, the restricted domain should be [tex]\([0, \infty)\)[/tex].
- The inverse function in this domain is [tex]\( f^{-1}(x) = \sqrt{x + a} \)[/tex].
Therefore, the correct statements are:
The restricted domain for [tex]\( f \)[/tex] is [tex]\([0, \infty)\)[/tex]. The inverse function is [tex]\( f^{-1}(x) = \sqrt{x + a}\)[/tex].
