what are the removable discontinuities of the following function?

\f(x) = \frac{x^2 - 36}{x^3 - 36x}\

$x = -6$, $x = 0$, and $x = 6$
$x = -6$ and $x = 6$
$x = 0$ and $x = -6$
$x = 0$ and $x = 6$

Question

what are the removable discontinuities of the following function?

\f(x) = \frac{x^2 - 36}{x^3 - 36x}\

$x = -6$, $x = 0$, and $x = 6$
$x = -6$ and $x = 6$
$x = 0$ and $x = -6$
$x = 0$ and $x = 6$

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Removable Discontinuities" ], "new_concepts": [], "current_concepts": [ "Rational Functions", "Removable Discontinuities", "Polynomial Factoring" ] } ## Factor the numerator and denominator $$ \begin{aligned} \text{Numerator: } & x^2 - 36 = (x - 6)(x + 6) \ \text{Denominator: } & x^3 - 36x = x(x^2 - 36) = x(x - 6)(x + 6) \end{aligned} $$ ## Identify the domain and discontinuities $$ \begin{aligned} \text{Domain: } & x eq 0, \quad x eq 6, \quad x eq -6 \end{aligned} $$ ## Determine removable discontinuities $$ \begin{aligned} f(x) &= \frac{(x - 6)(x + 6)}{x(x - 6)(x + 6)} = \frac{1}{x} \quad \text{for } x eq \pm 6, 0 \ \text{Removable: } & x = -6 \text{ and } x = 6 \end{aligned} $$

$x = -6$, $x = 0$, and $x = 6$

$x = -6$ and $x = 6$

$x = 0$ and $x = -6$

$x = 0$ and $x = 6$

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Calculus", "Removable Discontinuities" ] }

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QUESTION IMAGE

Question

Explanation:

Factor the numerator and denominator

Identify the domain and discontinuities

Determine removable discontinuities

Answer: