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Sagot :
Given the function [tex]\( f(x) = \frac{x^2 + 2x + 4}{x-2} \)[/tex] and its derivative [tex]\( f'(x) = \frac{x(x-4)}{(x-2)^2} \)[/tex], we need to determine the intervals where [tex]\( f \)[/tex] is decreasing. A function is decreasing where its derivative is less than zero, i.e., [tex]\( f'(x) < 0 \)[/tex].
The derivative [tex]\( f'(x) = \frac{x(x-4)}{(x-2)^2} \)[/tex] consists of two parts:
1. The numerator: [tex]\( x(x-4) \)[/tex]
2. The denominator: [tex]\( (x-2)^2 \)[/tex]
The denominator [tex]\((x-2)^2 \)[/tex] is always positive for all [tex]\( x \)[/tex] except [tex]\( x = 2 \)[/tex], where it is zero (but we cannot have this point in the intervals because [tex]\( f'(x) \)[/tex] is undefined here). Therefore, the sign of [tex]\( f'(x) \)[/tex] depends solely on the numerator [tex]\( x(x-4) \)[/tex].
We will analyze when the numerator [tex]\( x(x-4) \)[/tex] is less than zero:
- The product [tex]\( x(x-4) \)[/tex] is negative when [tex]\( x \)[/tex] and [tex]\( x-4 \)[/tex] have opposite signs.
To find where [tex]\( x(x-4) < 0 \)[/tex]:
1. Solve [tex]\( x(x-4) = 0 \)[/tex] to find critical points:
[tex]\[ x = 0 \quad \text{or} \quad x = 4 \][/tex]
2. Test intervals around these critical points:
- For [tex]\( x < 0 \)[/tex]:
Both factors [tex]\( x \)[/tex] and [tex]\( (x-4) \)[/tex] are negative, so their product is positive.
- For [tex]\( 0 < x < 4 \)[/tex]:
[tex]\( x \)[/tex] is positive and [tex]\( (x-4) \)[/tex] is negative, so their product is negative.
- For [tex]\( x > 4 \)[/tex]:
Both factors [tex]\( x \)[/tex] and [tex]\( (x-4) \)[/tex] are positive, so their product is positive.
Thus, [tex]\( x(x-4) \)[/tex] is negative for [tex]\( 0 < x < 4 \)[/tex].
However, we must exclude [tex]\( x = 2 \)[/tex] from the interval because [tex]\( (x-2)^2 \)[/tex] in the denominator would be zero, making [tex]\( f'(x) \)[/tex] undefined at this point.
Therefore, the intervals where [tex]\( f \)[/tex] is decreasing are:
[tex]\[ (0, 2) \cup (2, 4) \][/tex]
Conclusively, the correct answer is:
(C) [tex]\((0, 2) \cup (2, 4)\)[/tex]
The derivative [tex]\( f'(x) = \frac{x(x-4)}{(x-2)^2} \)[/tex] consists of two parts:
1. The numerator: [tex]\( x(x-4) \)[/tex]
2. The denominator: [tex]\( (x-2)^2 \)[/tex]
The denominator [tex]\((x-2)^2 \)[/tex] is always positive for all [tex]\( x \)[/tex] except [tex]\( x = 2 \)[/tex], where it is zero (but we cannot have this point in the intervals because [tex]\( f'(x) \)[/tex] is undefined here). Therefore, the sign of [tex]\( f'(x) \)[/tex] depends solely on the numerator [tex]\( x(x-4) \)[/tex].
We will analyze when the numerator [tex]\( x(x-4) \)[/tex] is less than zero:
- The product [tex]\( x(x-4) \)[/tex] is negative when [tex]\( x \)[/tex] and [tex]\( x-4 \)[/tex] have opposite signs.
To find where [tex]\( x(x-4) < 0 \)[/tex]:
1. Solve [tex]\( x(x-4) = 0 \)[/tex] to find critical points:
[tex]\[ x = 0 \quad \text{or} \quad x = 4 \][/tex]
2. Test intervals around these critical points:
- For [tex]\( x < 0 \)[/tex]:
Both factors [tex]\( x \)[/tex] and [tex]\( (x-4) \)[/tex] are negative, so their product is positive.
- For [tex]\( 0 < x < 4 \)[/tex]:
[tex]\( x \)[/tex] is positive and [tex]\( (x-4) \)[/tex] is negative, so their product is negative.
- For [tex]\( x > 4 \)[/tex]:
Both factors [tex]\( x \)[/tex] and [tex]\( (x-4) \)[/tex] are positive, so their product is positive.
Thus, [tex]\( x(x-4) \)[/tex] is negative for [tex]\( 0 < x < 4 \)[/tex].
However, we must exclude [tex]\( x = 2 \)[/tex] from the interval because [tex]\( (x-2)^2 \)[/tex] in the denominator would be zero, making [tex]\( f'(x) \)[/tex] undefined at this point.
Therefore, the intervals where [tex]\( f \)[/tex] is decreasing are:
[tex]\[ (0, 2) \cup (2, 4) \][/tex]
Conclusively, the correct answer is:
(C) [tex]\((0, 2) \cup (2, 4)\)[/tex]
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