To determine the range of the function [tex]\( g(x) = 3|x-1| - 1 \)[/tex], let's go through a detailed analysis:
1. Understanding the Absolute Value Expression:
[tex]\[
|x-1|
\][/tex]
The absolute value function [tex]\( |x-1| \)[/tex] outputs the distance of [tex]\( x \)[/tex] from 1 on the real number line. As a result, [tex]\( |x-1| \)[/tex] is always non-negative, that is:
[tex]\[
|x-1| \geq 0
\][/tex]
2. Multiplying by 3:
[tex]\[
3|x-1|
\][/tex]
Since [tex]\( |x-1| \)[/tex] is non-negative, multiplying it by 3 keeps it non-negative:
[tex]\[
3|x-1| \geq 0
\][/tex]
3. Subtracting 1:
[tex]\[
3|x-1| - 1
\][/tex]
Subtracting 1 from a non-negative quantity [tex]\( 3|x-1| \)[/tex], we get:
[tex]\[
3|x-1| - 1 \geq -1
\][/tex]
Hence, the expression [tex]\( 3|x-1| - 1 \)[/tex] will always be greater than or equal to [tex]\(-1\)[/tex].
4. Finding the Minimum Value:
- The smallest value of [tex]\( |x-1| \)[/tex] is 0, which occurs when [tex]\( x = 1 \)[/tex]. Plugging in [tex]\( x = 1 \)[/tex] into the function:
[tex]\[
g(1) = 3|1-1| - 1 = 3 \cdot 0 - 1 = -1
\][/tex]
So, the minimum value of [tex]\( g(x) \)[/tex] is [tex]\(-1\)[/tex].
5. Analyzing for Larger Values:
- As [tex]\( x \)[/tex] moves away from 1, [tex]\( |x-1| \)[/tex] increases. Consequently, [tex]\( 3|x-1| \)[/tex] becomes larger, and [tex]\( 3|x-1| - 1 \)[/tex] increases. Thus, as [tex]\( x \)[/tex] varies over all real numbers, [tex]\( g(x) \)[/tex] can take on any value starting from [tex]\(-1\)[/tex] and increasing without bound.
6. Final Range:
- Since [tex]\( 3|x-1| - 1 \)[/tex] can be -1 when [tex]\( x = 1 \)[/tex] and can increase indefinitely as [tex]\( x \)[/tex] moves away from 1, the range of [tex]\( g(x) \)[/tex] is:
[tex]\[
[-1, \infty)
\][/tex]
Given the choices:
A. [tex]\((- \infty, 1]\)[/tex] - Incorrect
B. [tex]\([-1, \infty)\)[/tex] - Correct
C. [tex]\([1, \infty)\)[/tex] - Incorrect
D. [tex]\((- \infty, \infty)\)[/tex] - Incorrect
The correct answer is:
[tex]\[
\boxed{B}
\][/tex]
