To find the interval where both functions [tex]\( f(x) = x^2 - x \)[/tex] and [tex]\( g(x) = \log(2x + 1) \)[/tex] are positive, we need to analyze each function separately and then find the common interval where both conditions are satisfied.
1. Determine where [tex]\( f(x) = x^2 - x > 0 \)[/tex]:
The quadratic [tex]\( x^2 - x \)[/tex] can be factored as:
[tex]\[
x^2 - x = x(x - 1)
\][/tex]
Therefore, [tex]\( x(x - 1) > 0 \)[/tex].
- The product of two numbers is positive when both numbers are positive or both numbers are negative.
- [tex]\( x(x - 1) > 0 \)[/tex] when [tex]\( x > 1 \)[/tex] or [tex]\( x < 0 \)[/tex] (excluding [tex]\( x = 0 \)[/tex] and [tex]\( x = 1 \)[/tex], since [tex]\( f(x) = 0 \)[/tex] at these points).
So, [tex]\( f(x) > 0 \)[/tex] for [tex]\( x \in (-\infty, 0) \cup (1, \infty) \)[/tex].
2. Determine where [tex]\( g(x) = \log(2x + 1) > 0 \)[/tex]:
The function [tex]\( \log(2x + 1) \)[/tex] is positive when the argument of the logarithm is greater than 1:
[tex]\[
2x + 1 > 1
\][/tex]
Solving the inequality:
[tex]\[
2x + 1 > 1 \implies 2x > 0 \implies x > 0
\][/tex]
So, [tex]\( g(x) > 0 \)[/tex] for [tex]\( x \in (0, \infty) \)[/tex].
3. Find the intersection of intervals:
We need the values of [tex]\( x \)[/tex] that satisfy both conditions:
[tex]\[
f(x) > 0 \quad \text{and} \quad g(x) > 0
\][/tex]
From the conditions above:
- [tex]\( f(x) > 0 \)[/tex] for [tex]\( x \in (-\infty, 0) \cup (1, \infty) \)[/tex]
- [tex]\( g(x) > 0 \)[/tex] for [tex]\( x \in (0, \infty) \)[/tex]
The intersection of these intervals where both functions are positive is:
[tex]\[
(1, \infty)
\][/tex]
Therefore, the correct interval where both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are positive is:
[tex]\((1, \infty)\)[/tex]
