Question

identify whether each value of \\(x\\) is a discontinuity of the function by typing asymptote, hole, or neither.

\\\frac{5x}{x^3 + 5x^2 + 6x}\\

\\(x = -3\\)
\\(x = -2\\)
\\(x = 0\\)
\\(x = 2\\)
\\(x = 3\\)
\\(x = 5\\)

Explanation:

Factor the denominator of the rational function

$$ x^3 + 5x^2 + 6x = x(x^2 + 5x + 6) = x(x + 2)(x + 3) $$

Rewrite the function and identify discontinuities

$$ f(x) = \frac{5x}{x(x + 2)(x + 3)} $$

The denominator is zero at \(x = 0\), \(x = -2\), and \(x = -3\).

• At \(x = 0\), the factor \(x\) is common to both numerator and denominator, creating a removable discontinuity (hole).
• At \(x = -2\) and \(x = -3\), the factors \((x+2)\) and \((x+3)\) do not cancel, creating non-removable discontinuities (vertical asymptotes).
• For \(x = 2\), \(x = 3\), and \(x = 5\), the denominator is non-zero, so there is no discontinuity (neither).

Evaluate each given value

• \(x = -3\): asymptote
• \(x = -2\): asymptote
• \(x = 0\): hole
• \(x = 2\): neither
• \(x = 3\): neither
• \(x = 5\): neither

Answer:

No.	Problem	Answer
2	\(x = -2\)	asymptote
3	\(x = 0\)	hole
4	\(x = 2\)	neither
5	\(x = 3\)	neither
6	\(x = 5\)	neither