Answer :
To determine the range of the piecewise function
[tex]\[ f(x) = \begin{cases} x^2 - 2 & \text{if } x < 0 \\ -3x - 2 & \text{if } x \geq 0 \end{cases} \][/tex]
we need to examine the behavior of the function across its different segments.
### 1. For [tex]\( x < 0 \)[/tex]: [tex]\( f(x) = x^2 - 2 \)[/tex]
In this segment, the function [tex]\( f(x) \)[/tex] is a parabola opening upward because the coefficient of [tex]\( x^2 \)[/tex] is positive. To understand the range for [tex]\( x < 0 \)[/tex]:
- Minimum value: As [tex]\( x \)[/tex] approaches 0 from the left (i.e., [tex]\( x \to 0^- \)[/tex]), [tex]\( x^2 - 2 \)[/tex] approaches [tex]\(-2\)[/tex]. Thus, the minimum value in this segment is [tex]\(-2\)[/tex].
- Maximum value: As [tex]\( x \)[/tex] decreases towards negative infinity (i.e., [tex]\( x \to -\infty \)[/tex]), [tex]\( x^2 - 2 \)[/tex] increases without bound because [tex]\( x^2 \)[/tex] becomes very large. Therefore, the maximum value in this segment is [tex]\( +\infty \)[/tex].
So, in the interval [tex]\( x < 0 \)[/tex], the range is [tex]\([-2, \infty)\)[/tex].
### 2. For [tex]\( x \geq 0 \)[/tex]: [tex]\( f(x) = -3x - 2 \)[/tex]
In this segment, the function [tex]\( f(x) \)[/tex] is a linear equation with a negative slope. To understand the range for [tex]\( x \geq 0 \)[/tex]:
- Maximum value: As [tex]\( x \)[/tex] approaches 0 from the right (i.e., [tex]\( x \to 0^+ \)[/tex]), [tex]\( -3x - 2 \)[/tex] approaches [tex]\(-2\)[/tex]. Thus, the maximum value in this segment is [tex]\(-2\)[/tex].
- Minimum value: As [tex]\( x \)[/tex] increases towards positive infinity (i.e., [tex]\( x \to +\infty \)[/tex]), [tex]\( -3x - 2 \)[/tex] decreases without bound. Therefore, the minimum value in this segment is [tex]\(-\infty\)[/tex].
So, in the interval [tex]\( x \geq 0 \)[/tex], the range is [tex]\((-\infty, -2]\)[/tex].
### Combining Both Segments
Given that there is no discontinuity at [tex]\( x = 0 \)[/tex] and the function smoothly transitions from [tex]\( x^2 - 2 \)[/tex] to [tex]\( -3x - 2 \)[/tex] at [tex]\( x = 0 \)[/tex], we can combine the ranges of the two segments.
- The minimum value of the entire function is [tex]\(-\infty\)[/tex] (from [tex]\( x \geq 0 \)[/tex] segment).
- The maximum value of the entire function is [tex]\( +\infty\)[/tex] (from [tex]\( x < 0 \)[/tex] segment).
Therefore, the range of the entire piecewise function is [tex]\( (-\infty, \infty) \)[/tex].
[tex]\[ f(x) = \begin{cases} x^2 - 2 & \text{if } x < 0 \\ -3x - 2 & \text{if } x \geq 0 \end{cases} \][/tex]
we need to examine the behavior of the function across its different segments.
### 1. For [tex]\( x < 0 \)[/tex]: [tex]\( f(x) = x^2 - 2 \)[/tex]
In this segment, the function [tex]\( f(x) \)[/tex] is a parabola opening upward because the coefficient of [tex]\( x^2 \)[/tex] is positive. To understand the range for [tex]\( x < 0 \)[/tex]:
- Minimum value: As [tex]\( x \)[/tex] approaches 0 from the left (i.e., [tex]\( x \to 0^- \)[/tex]), [tex]\( x^2 - 2 \)[/tex] approaches [tex]\(-2\)[/tex]. Thus, the minimum value in this segment is [tex]\(-2\)[/tex].
- Maximum value: As [tex]\( x \)[/tex] decreases towards negative infinity (i.e., [tex]\( x \to -\infty \)[/tex]), [tex]\( x^2 - 2 \)[/tex] increases without bound because [tex]\( x^2 \)[/tex] becomes very large. Therefore, the maximum value in this segment is [tex]\( +\infty \)[/tex].
So, in the interval [tex]\( x < 0 \)[/tex], the range is [tex]\([-2, \infty)\)[/tex].
### 2. For [tex]\( x \geq 0 \)[/tex]: [tex]\( f(x) = -3x - 2 \)[/tex]
In this segment, the function [tex]\( f(x) \)[/tex] is a linear equation with a negative slope. To understand the range for [tex]\( x \geq 0 \)[/tex]:
- Maximum value: As [tex]\( x \)[/tex] approaches 0 from the right (i.e., [tex]\( x \to 0^+ \)[/tex]), [tex]\( -3x - 2 \)[/tex] approaches [tex]\(-2\)[/tex]. Thus, the maximum value in this segment is [tex]\(-2\)[/tex].
- Minimum value: As [tex]\( x \)[/tex] increases towards positive infinity (i.e., [tex]\( x \to +\infty \)[/tex]), [tex]\( -3x - 2 \)[/tex] decreases without bound. Therefore, the minimum value in this segment is [tex]\(-\infty\)[/tex].
So, in the interval [tex]\( x \geq 0 \)[/tex], the range is [tex]\((-\infty, -2]\)[/tex].
### Combining Both Segments
Given that there is no discontinuity at [tex]\( x = 0 \)[/tex] and the function smoothly transitions from [tex]\( x^2 - 2 \)[/tex] to [tex]\( -3x - 2 \)[/tex] at [tex]\( x = 0 \)[/tex], we can combine the ranges of the two segments.
- The minimum value of the entire function is [tex]\(-\infty\)[/tex] (from [tex]\( x \geq 0 \)[/tex] segment).
- The maximum value of the entire function is [tex]\( +\infty\)[/tex] (from [tex]\( x < 0 \)[/tex] segment).
Therefore, the range of the entire piecewise function is [tex]\( (-\infty, \infty) \)[/tex].