Answer the question on a separate piece of paper. Show your work.

What is the range of [tex]$f(x)=\left\{\begin{array}{ll}x^2-1 & \text { for } x\ \textless \ 0 \\ 2x-1 & \text { for } 0 \leq x\ \textless \ 1 \\ x+2 & \text { for } x \geq 1\end{array}\right.$[/tex]?

(Note: This question will be marked incorrect until your facilitator grades the question using the provided answer. You may write "See paper" in the answer box.)



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.