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Sagot :
To determine whether each given value of [tex]\( x \)[/tex] is a discontinuity of the function [tex]\( f(x) = \frac{5x}{x^3 + 5x^2 + 6x} \)[/tex] as an asymptote, hole, or neither, we proceed as follows:
1. Factor the Denominator:
The denominator [tex]\( x^3 + 5x^2 + 6x \)[/tex] can be factorized:
[tex]\[ x^3 + 5x^2 + 6x = x(x^2 + 5x + 6) = x(x+2)(x+3) \][/tex]
Thus, the function can be rewritten as:
[tex]\[ f(x) = \frac{5x}{x(x+2)(x+3)} \][/tex]
2. Identify the Roots of the Denominator:
The roots of the denominator, which are the potential points of discontinuity, are:
[tex]\[ x = 0, \quad x = -2, \quad x = -3 \][/tex]
3. Examine Each Given Value of [tex]\( x \)[/tex]:
- For [tex]\( x = -3 \)[/tex]:
The denominator becomes zero, specifically at [tex]\((x+3)\)[/tex] term, suggesting a potential discontinuity.
Since the function can't be simplified to eliminate the [tex]\( (x+3) \)[/tex] term, this indicates an asymptote.
- For [tex]\( x = -2 \)[/tex]:
The denominator becomes zero, specifically at [tex]\((x+2)\)[/tex] term, implying a potential discontinuity.
Since the function can't be simplified to eliminate the [tex]\( (x+2) \)[/tex] term, this indicates an asymptote.
- For [tex]\( x = 0 \)[/tex]:
Both the numerator and the denominator become zero, suggesting a potential removable discontinuity.
Simplification shows [tex]\( \frac{5x}{x(x+2)(x+3)} = \frac{5}{(x+2)(x+3)} \)[/tex] when [tex]\((x \neq 0)\)[/tex].
This simplification confirms it is a hole, as the function is not defined at [tex]\( x = 0 \)[/tex], but it can be made continuous by defining the limit at [tex]\( x = 0 \)[/tex].
- For [tex]\( x = 2 \)[/tex]:
Substituting [tex]\( x = 2 \)[/tex] into the function doesn't make the denominator zero, implying no discontinuity at this point.
Hence, [tex]\( x = 2 \)[/tex] is neither an asymptote nor a hole.
- For [tex]\( x = 3 \)[/tex]:
Similarly, substituting [tex]\( x = 3 \)[/tex] into the function doesn't make the denominator zero, thus no discontinuity at this point.
Therefore, [tex]\( x = 3 \)[/tex] is neither an asymptote nor a hole.
### Summary:
[tex]\[ \begin{array}{l} \frac{5 x}{x^3+5 x^2+6 x} \\ x=-3 \quad \text{asymptote} \\ x=-2 \quad \text{asymptote} \\ x=0 \quad \text{hole} \\ x=2 \quad \text{neither} \\ x=3 \quad \text{neither} \end{array} \][/tex]
1. Factor the Denominator:
The denominator [tex]\( x^3 + 5x^2 + 6x \)[/tex] can be factorized:
[tex]\[ x^3 + 5x^2 + 6x = x(x^2 + 5x + 6) = x(x+2)(x+3) \][/tex]
Thus, the function can be rewritten as:
[tex]\[ f(x) = \frac{5x}{x(x+2)(x+3)} \][/tex]
2. Identify the Roots of the Denominator:
The roots of the denominator, which are the potential points of discontinuity, are:
[tex]\[ x = 0, \quad x = -2, \quad x = -3 \][/tex]
3. Examine Each Given Value of [tex]\( x \)[/tex]:
- For [tex]\( x = -3 \)[/tex]:
The denominator becomes zero, specifically at [tex]\((x+3)\)[/tex] term, suggesting a potential discontinuity.
Since the function can't be simplified to eliminate the [tex]\( (x+3) \)[/tex] term, this indicates an asymptote.
- For [tex]\( x = -2 \)[/tex]:
The denominator becomes zero, specifically at [tex]\((x+2)\)[/tex] term, implying a potential discontinuity.
Since the function can't be simplified to eliminate the [tex]\( (x+2) \)[/tex] term, this indicates an asymptote.
- For [tex]\( x = 0 \)[/tex]:
Both the numerator and the denominator become zero, suggesting a potential removable discontinuity.
Simplification shows [tex]\( \frac{5x}{x(x+2)(x+3)} = \frac{5}{(x+2)(x+3)} \)[/tex] when [tex]\((x \neq 0)\)[/tex].
This simplification confirms it is a hole, as the function is not defined at [tex]\( x = 0 \)[/tex], but it can be made continuous by defining the limit at [tex]\( x = 0 \)[/tex].
- For [tex]\( x = 2 \)[/tex]:
Substituting [tex]\( x = 2 \)[/tex] into the function doesn't make the denominator zero, implying no discontinuity at this point.
Hence, [tex]\( x = 2 \)[/tex] is neither an asymptote nor a hole.
- For [tex]\( x = 3 \)[/tex]:
Similarly, substituting [tex]\( x = 3 \)[/tex] into the function doesn't make the denominator zero, thus no discontinuity at this point.
Therefore, [tex]\( x = 3 \)[/tex] is neither an asymptote nor a hole.
### Summary:
[tex]\[ \begin{array}{l} \frac{5 x}{x^3+5 x^2+6 x} \\ x=-3 \quad \text{asymptote} \\ x=-2 \quad \text{asymptote} \\ x=0 \quad \text{hole} \\ x=2 \quad \text{neither} \\ x=3 \quad \text{neither} \end{array} \][/tex]
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