Sure, let's break down the problem step by step to determine the range of the given piecewise function:
Given the function:
[tex]\[ f(x) = \begin{cases}
-x, & \text{if } x < a \\
-1, & \text{if } a \leq x < b \\
(x-3)^2 - 1, & \text{if } x \geq b
\end{cases} \][/tex]
where [tex]\( a = 1 \)[/tex].
We need to analyze each piece of the function to identify the range:
1. First piece: [tex]\( f(x) = -x \)[/tex] for [tex]\( x < a \)[/tex] (i.e., [tex]\( x < 1 \)[/tex])
- As [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex], [tex]\( -x \)[/tex] approaches [tex]\( \infty \)[/tex].
- As [tex]\( x \)[/tex] approaches 1 from the left, [tex]\( -x \)[/tex] approaches -1.
Therefore, the values of [tex]\( f(x) \)[/tex] in this interval can be [tex]\( (-\infty, -1) \)[/tex].
2. Second piece: [tex]\( f(x) = -1 \)[/tex] for [tex]\( a \leq x < b \)[/tex] (i.e., [tex]\( 1 \leq x < b \)[/tex])
- This piece of the function is a constant value of [tex]\(-1\)[/tex].
Since the function is constant and equal to -1 throughout this interval, [tex]\( f(x) \)[/tex] is equal to [tex]\(-1\)[/tex].
3. Third piece: [tex]\( f(x) = (x-3)^2 - 1 \)[/tex] for [tex]\( x \geq b \)[/tex]
- This is a quadratic function, specifically a parabola that opens upwards. The vertex is at [tex]\(x = 3\)[/tex], and the vertex value is [tex]\( (3-3)^2 - 1 = -1 \)[/tex].
- As [tex]\( x \)[/tex] increases or decreases away from 3, the value of [tex]\( (x-3)^2 \)[/tex] increases, making [tex]\( (x-3)^2 - 1 \)[/tex] greater than [tex]\(-1\)[/tex].
Therefore, the values of [tex]\( f(x) \)[/tex] in this interval can be [tex]\( [-1, \infty) \)[/tex].
Combining all these pieces:
- From the first piece, values range from [tex]\((- \infty, -1)\)[/tex].
- From the second piece, we have exactly [tex]\(-1\)[/tex].
- From the third piece, values are from [tex]\([-1, \infty)\)[/tex].
Putting it all together, the overall range of the function [tex]\( f(x) \)[/tex] is:
[tex]\[ (-1, \infty) \][/tex]
Thus, the range of the given function is:
[tex]\[ (-1, \infty) \][/tex]
The correct answer is:
[tex]\[ (-1, \infty) \][/tex]
