To determine the interval where both functions [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are positive, we need to analyze each function individually and then see where they overlap in terms of positiveness.
### Function [tex]\( f(x) \)[/tex]
Given:
- [tex]\( f(x) \)[/tex] is a logarithmic function with a vertical asymptote at [tex]\( x = 0 \)[/tex].
- [tex]\( f(x) \)[/tex] has an [tex]\( x \)[/tex]-intercept at [tex]\( (4, 0) \)[/tex].
- [tex]\( f(x) \)[/tex] is decreasing over the interval [tex]\( (0, \infty) \)[/tex].
From these properties, we can infer:
- Because [tex]\( f(x) \)[/tex] is logarithmic and has an intercept at [tex]\( (4, 0) \)[/tex], it must be in the form [tex]\( f(x) = \log_b(x) + c \)[/tex], where [tex]\( b > 1 \)[/tex] since the function is decreasing.
- Therefore, [tex]\( f(x) \)[/tex] will be positive when [tex]\( \log_b(x) + c > 0 \)[/tex].
Since the [tex]\( x \)[/tex]-intercept is at [tex]\( 4 \)[/tex]:
- For [tex]\( f(x) = \log_b(x) - \log_b(4) \)[/tex]:
- [tex]\( f(x) = \log_b(x/4) \)[/tex]
We need [tex]\( x \)[/tex] values such that [tex]\( x / 4 > 1 \)[/tex], i.e., [tex]\( x > 4 \)[/tex].
### Function [tex]\( g(x) \)[/tex]
Given:
- [tex]\( g(x) = \log_2(x + 3) - 2 \)[/tex]
For [tex]\( g(x) \)[/tex] to be positive:
- [tex]\( \log_2(x + 3) - 2 > 0 \)[/tex]
- [tex]\( \log_2(x + 3) > 2 \)[/tex]
- This implies that [tex]\( x + 3 > 2^2 \)[/tex]
- [tex]\( x + 3 > 4 \)[/tex]
- [tex]\( x > 1 \)[/tex]
### Intersection of Positive Intervals
Now, combining the intervals where each function is positive:
- [tex]\( f(x) \)[/tex] is positive for [tex]\( x > 4 \)[/tex]
- [tex]\( g(x) \)[/tex] is positive for [tex]\( x > 1 \)[/tex]
The overlap of these intervals is [tex]\( x > 4 \)[/tex].
### Conclusion
Both functions [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are positive for [tex]\( x \)[/tex] in the interval [tex]\( (4, \infty) \)[/tex].
