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:

⚡ Using what you learned: Types of Discontinuity (removable, jump, infinite)

Step 1: Factor the denominator

Factor the denominator of the function \( f(x) = \frac{5x}{x^3 + 5x^2 + 6x} \):

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

The function can be written as:

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

Step 2: Identify discontinuities and their types

The function is undefined where the denominator is zero: \( x = 0 \), \( x = -2 \), and \( x = -3 \).

• For \( x = 0 \): The factor \( x \) is common to both the numerator and the denominator. It can be canceled out (removable discontinuity), which represents a hole.
• For \( x = -2 \): The factor \( x + 2 \) remains in the denominator after simplification, causing the function to approach infinity (non-removable infinite discontinuity), which represents an asymptote.
• For \( x = -3 \): The factor \( x + 3 \) remains in the denominator after simplification, which represents an asymptote.
• For \( x = 2 \), \( x = 3 \), and \( x = 5 \): The denominator is not zero at these values, so there is no discontinuity (neither).

Answer:

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