Answer :
To determine the range of the piecewise-defined function [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = \begin{cases} x^2 - 1 & \text{for } x < 0 \\ 2x - 1 & \text{for } 0 \leq x < 1 \\ x + 2 & \text{for } x \geq 1 \end{cases} \][/tex]
let's analyze each piece at a time.
### 1. First Piece: [tex]\( f(x) = x^2 - 1 \)[/tex] for [tex]\( x < 0 \)[/tex]
For [tex]\( x < 0 \)[/tex], the function is a downward-facing parabola shifted down by 1 unit. As [tex]\( x \)[/tex] approaches 0 from the left:
- When [tex]\( x \)[/tex] is close to 0 but negative, [tex]\( x^2 \)[/tex] is close to 0. Therefore, [tex]\( x^2 - 1 \)[/tex] is close to -1.
- As [tex]\( x \)[/tex] becomes very large in the negative direction (negatively unbounded), [tex]\( x^2 \)[/tex] becomes very large, making [tex]\( x^2 - 1 \)[/tex] also very large.
Thus, for [tex]\( x < 0 \)[/tex], [tex]\( f(x) \)[/tex] ranges from [tex]\(-1\)[/tex] (approached as [tex]\( x \)[/tex] nears 0 from the negative side) to [tex]\( \infty \)[/tex].
### 2. Second Piece: [tex]\( f(x) = 2x - 1 \)[/tex] for [tex]\( 0 \leq x < 1 \)[/tex]
For [tex]\( 0 \leq x < 1 \)[/tex], the function is linear with a slope of 2. Evaluating at the boundaries:
- When [tex]\( x = 0 \)[/tex], [tex]\( f(x) = 2(0) - 1 = -1 \)[/tex].
- When [tex]\( x \)[/tex] is close to but less than 1, [tex]\( f(x) = 2(1) - 1 = 1 \)[/tex].
Thus, for [tex]\( 0 \leq x < 1 \)[/tex], [tex]\( f(x) \)[/tex] ranges from [tex]\( -1 \)[/tex] to just under [tex]\( 1 \)[/tex] (not including 1).
### 3. Third Piece: [tex]\( f(x) = x + 2 \)[/tex] for [tex]\( x \geq 1 \)[/tex]
For [tex]\( x \geq 1 \)[/tex], the function is linear with a slope of 1. Evaluating at the boundaries:
- When [tex]\( x = 1 \)[/tex], [tex]\( f(x) = 1 + 2 = 3 \)[/tex].
Since there is no upper bound for [tex]\( x \)[/tex]:
- As [tex]\( x \)[/tex] increases, [tex]\( f(x) = x + 2 \)[/tex] keeps increasing.
Thus, for [tex]\( x \geq 1 \)[/tex], [tex]\( f(x) \)[/tex] ranges from [tex]\( 3 \)[/tex] to [tex]\( \infty \)[/tex].
### Combining the Ranges
Considering all three pieces together:
- The first piece ranges from [tex]\(-1\)[/tex] to [tex]\(\infty\)[/tex].
- The second piece ranges from [tex]\(-1\)[/tex] to just below [tex]\(1\)[/tex].
- The third piece starts at [tex]\(3\)[/tex] and goes to [tex]\(\infty\)[/tex].
Combining these, the overall range of [tex]\( f(x) \)[/tex] is:
[tex]\[ \boxed{[-1, \infty)} \][/tex]
This range includes all values from [tex]\(-1\)[/tex] onwards, without any gaps in between.
[tex]\[ f(x) = \begin{cases} x^2 - 1 & \text{for } x < 0 \\ 2x - 1 & \text{for } 0 \leq x < 1 \\ x + 2 & \text{for } x \geq 1 \end{cases} \][/tex]
let's analyze each piece at a time.
### 1. First Piece: [tex]\( f(x) = x^2 - 1 \)[/tex] for [tex]\( x < 0 \)[/tex]
For [tex]\( x < 0 \)[/tex], the function is a downward-facing parabola shifted down by 1 unit. As [tex]\( x \)[/tex] approaches 0 from the left:
- When [tex]\( x \)[/tex] is close to 0 but negative, [tex]\( x^2 \)[/tex] is close to 0. Therefore, [tex]\( x^2 - 1 \)[/tex] is close to -1.
- As [tex]\( x \)[/tex] becomes very large in the negative direction (negatively unbounded), [tex]\( x^2 \)[/tex] becomes very large, making [tex]\( x^2 - 1 \)[/tex] also very large.
Thus, for [tex]\( x < 0 \)[/tex], [tex]\( f(x) \)[/tex] ranges from [tex]\(-1\)[/tex] (approached as [tex]\( x \)[/tex] nears 0 from the negative side) to [tex]\( \infty \)[/tex].
### 2. Second Piece: [tex]\( f(x) = 2x - 1 \)[/tex] for [tex]\( 0 \leq x < 1 \)[/tex]
For [tex]\( 0 \leq x < 1 \)[/tex], the function is linear with a slope of 2. Evaluating at the boundaries:
- When [tex]\( x = 0 \)[/tex], [tex]\( f(x) = 2(0) - 1 = -1 \)[/tex].
- When [tex]\( x \)[/tex] is close to but less than 1, [tex]\( f(x) = 2(1) - 1 = 1 \)[/tex].
Thus, for [tex]\( 0 \leq x < 1 \)[/tex], [tex]\( f(x) \)[/tex] ranges from [tex]\( -1 \)[/tex] to just under [tex]\( 1 \)[/tex] (not including 1).
### 3. Third Piece: [tex]\( f(x) = x + 2 \)[/tex] for [tex]\( x \geq 1 \)[/tex]
For [tex]\( x \geq 1 \)[/tex], the function is linear with a slope of 1. Evaluating at the boundaries:
- When [tex]\( x = 1 \)[/tex], [tex]\( f(x) = 1 + 2 = 3 \)[/tex].
Since there is no upper bound for [tex]\( x \)[/tex]:
- As [tex]\( x \)[/tex] increases, [tex]\( f(x) = x + 2 \)[/tex] keeps increasing.
Thus, for [tex]\( x \geq 1 \)[/tex], [tex]\( f(x) \)[/tex] ranges from [tex]\( 3 \)[/tex] to [tex]\( \infty \)[/tex].
### Combining the Ranges
Considering all three pieces together:
- The first piece ranges from [tex]\(-1\)[/tex] to [tex]\(\infty\)[/tex].
- The second piece ranges from [tex]\(-1\)[/tex] to just below [tex]\(1\)[/tex].
- The third piece starts at [tex]\(3\)[/tex] and goes to [tex]\(\infty\)[/tex].
Combining these, the overall range of [tex]\( f(x) \)[/tex] is:
[tex]\[ \boxed{[-1, \infty)} \][/tex]
This range includes all values from [tex]\(-1\)[/tex] onwards, without any gaps in between.