To determine the intervals for which the value of [tex]\( (f-g)(x) \)[/tex] is negative, we need to proceed step-by-step by analyzing the behavior of the functions [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] on each given interval. Let's denote [tex]\( f(x) = 3x + 5 \)[/tex] and [tex]\( g(x) = x + 7 \)[/tex].
The function [tex]\( (f-g)(x) \)[/tex] is defined as:
[tex]\[ (f-g)(x) = f(x) - g(x) = (3x + 5) - (x + 7) = 3x + 5 - x - 7 = 2x - 2 \][/tex]
We need to determine the intervals where [tex]\( (f-g)(x) < 0 \)[/tex].
This can be expressed as:
[tex]\[ 2x - 2 < 0 \][/tex]
Solving this inequality:
[tex]\[ 2x < 2 \][/tex]
[tex]\[ x < 1 \][/tex]
Our goal now is to check each given interval to see if it fits within [tex]\( x < 1 \)[/tex]:
1. For the interval [tex]\( (-\infty, -1) \)[/tex]:
- All values [tex]\( x \)[/tex] in this interval are less than 1. Therefore, [tex]\( 2x - 2 < 0 \)[/tex] for all [tex]\( x \in (-\infty, -1) \)[/tex].
2. For the interval [tex]\( (-\infty, 2) \)[/tex]:
- This interval includes all values less than 2, so it does include the values where [tex]\( x < 1 \)[/tex], making [tex]\( 2x - 2 < 0 \)[/tex] for [tex]\( x < 1 \)[/tex]. However, it also includes [tex]\( x \)[/tex] values between 1 and 2, where [tex]\( 2x - 2 \)[/tex] is not less than 0.
3. For the interval [tex]\( (0, 3) \)[/tex]:
- This interval includes some values less than 1 (0 to 1), where [tex]\( 2x - 2 < 0 \)[/tex], but it extends to 3, where not all values make [tex]\( 2x - 2 \)[/tex] negative.
4. For the interval [tex]\( (2, \infty) \)[/tex]:
- All values [tex]\( x \)[/tex] in this interval are greater than 2, hence [tex]\( 2x - 2 \)[/tex] is positive in this entire interval and doesn't satisfy [tex]\( (f-g)(x) < 0 \)[/tex].
After analyzing the above intervals with respect to [tex]\( x < 1 \)[/tex], the interval which solely makes [tex]\( (f-g)(x) \)[/tex] negative is:
[tex]\[ (-\infty, -1) \][/tex]
Therefore, the value of [tex]\( (f-g)(x) \)[/tex] is negative in the interval [tex]\( (-\infty, -1) \)[/tex]. However, based on the final result, we observe that there are no intervals where [tex]\( (f-g)(x) < 0 \)[/tex] overall, indicated by the empty result. Thus, there isn't a single interval where [tex]\( (f-g)(x) \)[/tex] remains consistently negative across any point.
