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Sagot :
Certainly! Let's analyze the functions [tex]\( f(x) = x^2 - x \)[/tex] and [tex]\( g(x) = \log(2x + 1) \)[/tex], and determine on which interval both functions are positive.
Firstly, let's examine [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = x^2 - x \][/tex]
To find where [tex]\( f(x) \)[/tex] is positive, we need to solve the inequality:
[tex]\[ x^2 - x > 0 \][/tex]
[tex]\[ x(x - 1) > 0 \][/tex]
This quadratic inequality factors into the product of two binomials. The solutions to [tex]\( x(x - 1) = 0 \)[/tex] are [tex]\( x = 0 \)[/tex] and [tex]\( x = 1 \)[/tex]. To determine where the product is positive, we analyze the intervals defined by these critical points:
- For [tex]\( x < 0 \)[/tex], both [tex]\( x \)[/tex] and [tex]\( x-1 \)[/tex] are negative, so their product is positive.
- For [tex]\( 0 < x < 1 \)[/tex], [tex]\( x \)[/tex] is positive and [tex]\( x-1 \)[/tex] is negative, so the product is negative.
- For [tex]\( x > 1 \)[/tex], both [tex]\( x \)[/tex] and [tex]\( x-1 \)[/tex] are positive, so their product is positive.
Therefore, [tex]\( f(x) > 0 \)[/tex] for [tex]\( x \in (-\infty, 0) \cup (1, \infty) \)[/tex].
Next, let's examine [tex]\( g(x) \)[/tex]:
[tex]\[ g(x) = \log(2x + 1) \][/tex]
To find where [tex]\( g(x) \)[/tex] is positive, we need to solve the inequality:
[tex]\[ \log(2x + 1) > 0 \][/tex]
Since the natural logarithm [tex]\(\log(y)\)[/tex] is positive when [tex]\( y > 1 \)[/tex]:
[tex]\[ 2x + 1 > 1 \][/tex]
[tex]\[ 2x > 0 \][/tex]
[tex]\[ x > 0 \][/tex]
Therefore, [tex]\( g(x) > 0 \)[/tex] for [tex]\( x \in (0, \infty) \)[/tex].
Now let's combine these intervals to determine where both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are positive simultaneously:
- For [tex]\( (-\infty, 0) \)[/tex], [tex]\( g(x) \)[/tex] is not positive because [tex]\( g(x) \)[/tex] is positive only for [tex]\( x > 0 \)[/tex].
- For [tex]\( (0, 1) \)[/tex], [tex]\( f(x) \)[/tex] is not positive because we found [tex]\( f(x) \)[/tex] to be negative in this interval.
- For [tex]\( (1, \infty) \)[/tex], [tex]\( f(x) \)[/tex] is positive and [tex]\( g(x) \)[/tex] is positive.
After thorough analysis, we conclude that there is no common interval where both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are positive.
Thus, the result shows that there is no interval where both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are positive simultaneously.
Firstly, let's examine [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = x^2 - x \][/tex]
To find where [tex]\( f(x) \)[/tex] is positive, we need to solve the inequality:
[tex]\[ x^2 - x > 0 \][/tex]
[tex]\[ x(x - 1) > 0 \][/tex]
This quadratic inequality factors into the product of two binomials. The solutions to [tex]\( x(x - 1) = 0 \)[/tex] are [tex]\( x = 0 \)[/tex] and [tex]\( x = 1 \)[/tex]. To determine where the product is positive, we analyze the intervals defined by these critical points:
- For [tex]\( x < 0 \)[/tex], both [tex]\( x \)[/tex] and [tex]\( x-1 \)[/tex] are negative, so their product is positive.
- For [tex]\( 0 < x < 1 \)[/tex], [tex]\( x \)[/tex] is positive and [tex]\( x-1 \)[/tex] is negative, so the product is negative.
- For [tex]\( x > 1 \)[/tex], both [tex]\( x \)[/tex] and [tex]\( x-1 \)[/tex] are positive, so their product is positive.
Therefore, [tex]\( f(x) > 0 \)[/tex] for [tex]\( x \in (-\infty, 0) \cup (1, \infty) \)[/tex].
Next, let's examine [tex]\( g(x) \)[/tex]:
[tex]\[ g(x) = \log(2x + 1) \][/tex]
To find where [tex]\( g(x) \)[/tex] is positive, we need to solve the inequality:
[tex]\[ \log(2x + 1) > 0 \][/tex]
Since the natural logarithm [tex]\(\log(y)\)[/tex] is positive when [tex]\( y > 1 \)[/tex]:
[tex]\[ 2x + 1 > 1 \][/tex]
[tex]\[ 2x > 0 \][/tex]
[tex]\[ x > 0 \][/tex]
Therefore, [tex]\( g(x) > 0 \)[/tex] for [tex]\( x \in (0, \infty) \)[/tex].
Now let's combine these intervals to determine where both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are positive simultaneously:
- For [tex]\( (-\infty, 0) \)[/tex], [tex]\( g(x) \)[/tex] is not positive because [tex]\( g(x) \)[/tex] is positive only for [tex]\( x > 0 \)[/tex].
- For [tex]\( (0, 1) \)[/tex], [tex]\( f(x) \)[/tex] is not positive because we found [tex]\( f(x) \)[/tex] to be negative in this interval.
- For [tex]\( (1, \infty) \)[/tex], [tex]\( f(x) \)[/tex] is positive and [tex]\( g(x) \)[/tex] is positive.
After thorough analysis, we conclude that there is no common interval where both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are positive.
Thus, the result shows that there is no interval where both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are positive simultaneously.
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