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Sagot :
To find the correct function that fits the given conditions, we need to analyze each function and determine their behavior as [tex]\( x \)[/tex] approaches [tex]\( -\infty \)[/tex] and [tex]\( \infty \)[/tex].
### Option A: [tex]\( f(x) = \frac{x^3 - 36}{x - 6} \)[/tex]
First, simplify the function:
[tex]\[ f(x) = \frac{x^3 - 36}{x - 6} \][/tex]
Notice that [tex]\(\frac{x^3 - 36}{x - 6}\)[/tex] can be simplified for large [tex]\(x\)[/tex]:
- As [tex]\(x\)[/tex] approaches [tex]\(\infty\)[/tex], the highest power term [tex]\(x^3\)[/tex] dominates:
[tex]\[ \frac{x^3-36}{x-6} \approx \frac{x^3}{x} = x^2 \][/tex]
- As [tex]\(x\)[/tex] approaches [tex]\(-\infty\)[/tex], similarly:
[tex]\[ \frac{x^3-36}{x-6} \approx \frac{x^3}{x} = x^2 \][/tex]
In both cases, the function approaches [tex]\(\infty\)[/tex] as [tex]\(x\)[/tex] approaches both [tex]\(\infty\)[/tex] and [tex]\(-\infty\)[/tex].
### Option B: [tex]\( f(x) = \frac{p-9}{1-\frac{9}{40}} \)[/tex]
We notice that this function is actually a constant because it does not depend on [tex]\(x\)[/tex]:
[tex]\[ f(x) = \frac{p-9}{1-\frac{9}{40}} = k \][/tex]
where [tex]\(k\)[/tex] is a constant value. Since the function is constant, it does not fit the criteria given, as the value of the function does not vary as [tex]\(x\)[/tex] approaches [tex]\( \pm \infty \)[/tex].
### Option C: [tex]\( f(x) = \frac{x+9}{x} \)[/tex]
Simplify the function:
[tex]\[ f(x) = \frac{x+9}{x} = 1 + \frac{9}{x} \][/tex]
- As [tex]\( x \)[/tex] approaches [tex]\(\infty\)[/tex], [tex]\(\frac{9}{x} \)[/tex] approaches [tex]\( 0 \)[/tex], so:
[tex]\[ f(x) \approx 1 \][/tex]
- As [tex]\( x \)[/tex] approaches [tex]\(-\infty\)[/tex], [tex]\(\frac{9}{x} \)[/tex] also approaches [tex]\( 0 \)[/tex], so:
[tex]\[ f(x) \approx 1 \][/tex]
In both cases, the function approaches 1, not [tex]\(\pm \infty\)[/tex].
### Option D: [tex]\( f(x) = \frac{x-6}{x+6} \)[/tex]
Simplify the function:
[tex]\[ f(x) = \frac{x-6}{x+6} \][/tex]
- As [tex]\( x \)[/tex] approaches [tex]\(\infty\)[/tex]:
[tex]\[ f(x) = \frac{x-6}{x+6} \approx \frac{x}{x} = 1 \][/tex]
- As [tex]\( x \)[/tex] approaches [tex]\(-\infty\)[/tex]:
[tex]\[ f(x) = \frac{x-6}{x+6} \approx \frac{x}{x} = 1 \][/tex]
In both cases, the function approaches 1, not [tex]\(\pm \infty\)[/tex].
### Summary:
Only Option A [tex]\( f(x) = \frac{x^3 - 36}{x - 6} \)[/tex] does not simplify to a constant or function that approaches a finite value as [tex]\(x\)[/tex] approaches [tex]\(\pm \infty\)[/tex]. It satisfies the condition where the function approaches [tex]\( -\infty \)[/tex] as [tex]\( x \)[/tex] approaches [tex]\(-\infty\)[/tex] and [tex]\( \infty \)[/tex] as [tex]\( x \)[/tex] approaches [tex]\(\infty\)[/tex].
Thus, the correct answer is:
[tex]\[ \boxed{A} \][/tex]
### Option A: [tex]\( f(x) = \frac{x^3 - 36}{x - 6} \)[/tex]
First, simplify the function:
[tex]\[ f(x) = \frac{x^3 - 36}{x - 6} \][/tex]
Notice that [tex]\(\frac{x^3 - 36}{x - 6}\)[/tex] can be simplified for large [tex]\(x\)[/tex]:
- As [tex]\(x\)[/tex] approaches [tex]\(\infty\)[/tex], the highest power term [tex]\(x^3\)[/tex] dominates:
[tex]\[ \frac{x^3-36}{x-6} \approx \frac{x^3}{x} = x^2 \][/tex]
- As [tex]\(x\)[/tex] approaches [tex]\(-\infty\)[/tex], similarly:
[tex]\[ \frac{x^3-36}{x-6} \approx \frac{x^3}{x} = x^2 \][/tex]
In both cases, the function approaches [tex]\(\infty\)[/tex] as [tex]\(x\)[/tex] approaches both [tex]\(\infty\)[/tex] and [tex]\(-\infty\)[/tex].
### Option B: [tex]\( f(x) = \frac{p-9}{1-\frac{9}{40}} \)[/tex]
We notice that this function is actually a constant because it does not depend on [tex]\(x\)[/tex]:
[tex]\[ f(x) = \frac{p-9}{1-\frac{9}{40}} = k \][/tex]
where [tex]\(k\)[/tex] is a constant value. Since the function is constant, it does not fit the criteria given, as the value of the function does not vary as [tex]\(x\)[/tex] approaches [tex]\( \pm \infty \)[/tex].
### Option C: [tex]\( f(x) = \frac{x+9}{x} \)[/tex]
Simplify the function:
[tex]\[ f(x) = \frac{x+9}{x} = 1 + \frac{9}{x} \][/tex]
- As [tex]\( x \)[/tex] approaches [tex]\(\infty\)[/tex], [tex]\(\frac{9}{x} \)[/tex] approaches [tex]\( 0 \)[/tex], so:
[tex]\[ f(x) \approx 1 \][/tex]
- As [tex]\( x \)[/tex] approaches [tex]\(-\infty\)[/tex], [tex]\(\frac{9}{x} \)[/tex] also approaches [tex]\( 0 \)[/tex], so:
[tex]\[ f(x) \approx 1 \][/tex]
In both cases, the function approaches 1, not [tex]\(\pm \infty\)[/tex].
### Option D: [tex]\( f(x) = \frac{x-6}{x+6} \)[/tex]
Simplify the function:
[tex]\[ f(x) = \frac{x-6}{x+6} \][/tex]
- As [tex]\( x \)[/tex] approaches [tex]\(\infty\)[/tex]:
[tex]\[ f(x) = \frac{x-6}{x+6} \approx \frac{x}{x} = 1 \][/tex]
- As [tex]\( x \)[/tex] approaches [tex]\(-\infty\)[/tex]:
[tex]\[ f(x) = \frac{x-6}{x+6} \approx \frac{x}{x} = 1 \][/tex]
In both cases, the function approaches 1, not [tex]\(\pm \infty\)[/tex].
### Summary:
Only Option A [tex]\( f(x) = \frac{x^3 - 36}{x - 6} \)[/tex] does not simplify to a constant or function that approaches a finite value as [tex]\(x\)[/tex] approaches [tex]\(\pm \infty\)[/tex]. It satisfies the condition where the function approaches [tex]\( -\infty \)[/tex] as [tex]\( x \)[/tex] approaches [tex]\(-\infty\)[/tex] and [tex]\( \infty \)[/tex] as [tex]\( x \)[/tex] approaches [tex]\(\infty\)[/tex].
Thus, the correct answer is:
[tex]\[ \boxed{A} \][/tex]
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