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:

Step1: Factor the denominator

First, factor the denominator \(x^3 + 5x^2 + 6x\). We can factor out an \(x\) first: \(x(x^2 + 5x + 6)\). Then factor the quadratic: \(x(x + 2)(x + 3)\). So the function becomes \(\frac{5x}{x(x + 2)(x + 3)}\).

Step2: Simplify the function (cancel common factors)

We can cancel the common factor of \(x\) (for \(x
eq0\)), so the simplified function is \(\frac{5}{(x + 2)(x + 3)}\), with the restriction \(x
eq0, x
eq - 2, x
eq - 3\) (since these values make the original denominator zero).

Step3: Analyze each \(x\)-value

• For \(x=-3\): The original denominator is zero (since \((-3)+3 = 0\)) and the simplified denominator is also zero (because \((-3 + 2)(-3+3)=(-1)(0) = 0\)), but the numerator at \(x = - 3\) in the original function: numerator is \(5\times(-3)=-15\), denominator is \(0\), and since the factor \((x + 3)\) is in the simplified denominator (not canceled), this is a vertical asymptote.
• For \(x=-2\): Original denominator is zero (\((-2)+2 = 0\)) and simplified denominator is zero (\((-2 + 2)(-2+3)=(0)(1)=0\)), numerator at \(x=-2\) is \(5\times(-2)=-10\), denominator is \(0\), and the factor \((x + 2)\) is in the simplified denominator (not canceled), so this is a vertical asymptote.
• For \(x = 0\): Original denominator is zero (because \(x = 0\)) but we canceled the \(x\) factor. So the original function has a hole at \(x = 0\) (since the factor \(x\) was canceled, meaning the discontinuity is removable).
• For \(x=2\): Plug into the simplified function: denominator is \((2 + 2)(2+3)=(4)(5)=20

eq0\), numerator is \(5\), so the function is defined at \(x = 2\) (value is \(\frac{5}{20}=\frac{1}{4}\)), so neither discontinuity.

• For \(x=3\): Plug into the simplified function: denominator is \((3 + 2)(3+3)=(5)(6)=30

eq0\), numerator is \(5\), so the function is defined at \(x = 3\) (value is \(\frac{5}{30}=\frac{1}{6}\)), so neither discontinuity.

• For \(x=5\): Plug into the simplified function: denominator is \((5 + 2)(5+3)=(7)(8)=56

eq0\), numerator is \(5\), so the function is defined at \(x = 5\) (value is \(\frac{5}{56}\)), so neither discontinuity.

Answer:

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