Question

identify whether each value of x is a discontinuity of the function by choosing asymptote, hole, or neither. \\( \frac{5x}{x^3 + 5x^2 + 6x} \\) \\( x = -3 \\) \\( x = -2 \\) \\( x = 0 \\) \\( x = 2 \\) \\( x = 3 \\) \\( x = 5 \\)

Explanation:

First, we factor the denominator of the function \( f(x) = \frac{5x}{x^3 + 5x^2 + 6x} \).

Step 1: Factor the Denominator

Factor out an \( x \) from the denominator:
\( x^3 + 5x^2 + 6x = x(x^2 + 5x + 6) \)
Then factor the quadratic:
\( x^2 + 5x + 6 = (x + 2)(x + 3) \)
So the denominator becomes \( x(x + 2)(x + 3) \), and the function is \( f(x) = \frac{5x}{x(x + 2)(x + 3)} \), with \( x
eq 0, -2, -3 \) (since these make the denominator zero).

Step 2: Simplify the Function

Cancel out the common factor of \( x \) (for \( x
eq 0 \)):
\( f(x) = \frac{5}{(x + 2)(x + 3)} \), \( x
eq 0, -2, -3 \)

Now we analyze each value of \( x \):

For \( x = -3 \):

The original denominator is zero (since \( (-3)(-3 + 2)(-3 + 3) = (-3)(-1)(0) = 0 \)), and the simplified function still has a denominator of zero at \( x = -3 \) (since \( (-3 + 2)(-3 + 3) = (-1)(0) = 0 \)). So \( x = -3 \) is a vertical asymptote.

For \( x = -2 \):

The original denominator is zero ( \( (-2)(-2 + 2)(-2 + 3) = (-2)(0)(1) = 0 \) ), and the simplified function has a denominator of zero at \( x = -2 \) ( \( (-2 + 2)(-2 + 3) = (0)(1) = 0 \) ). So \( x = -2 \) is a vertical asymptote.

For \( x = 0 \):

The original denominator is zero, but we canceled out the \( x \) term. This means there is a hole at \( x = 0 \) (since the factor \( x \) was present in both numerator and denominator, so it's a removable discontinuity).

For \( x = 2 \):

Plug \( x = 2 \) into the simplified function: \( \frac{5}{(2 + 2)(2 + 3)} = \frac{5}{(4)(5)} = \frac{5}{20} = \frac{1}{4} \). The function is defined here, so neither.

For \( x = 3 \):

Plug \( x = 3 \) into the simplified function: \( \frac{5}{(3 + 2)(3 + 3)} = \frac{5}{(5)(6)} = \frac{5}{30} = \frac{1}{6} \). The function is defined here, so neither.

For \( x = 5 \):

Plug \( x = 5 \) into the simplified function: \( \frac{5}{(5 + 2)(5 + 3)} = \frac{5}{(7)(8)} = \frac{5}{56} \). The function is defined here, so neither.

Answer:

• \( x = -3 \): Asymptote
• \( x = -2 \): Asymptote
• \( x = 0 \): Hole
• \( x = 2 \): Neither
• \( x = 3 \): Neither
• \( x = 5 \): Neither