identify whether each value of x is a discontinuity of the function by typing asymptote, hole, or…

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

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

Answer

Explanation:

Step1: Factor the denominator

First, factor $x^{3}+5x^{2}+6x=x(x^{2}+5x + 6)=x(x + 2)(x+3)$. So the function is $f(x)=\frac{5x}{x(x + 2)(x + 3)}$.

Step2: Simplify the function

Cancel out the common factor $x$ (for $x\neq0$), we get $f(x)=\frac{5}{(x + 2)(x + 3)}$ for $x\neq0$.

Step3: Analyze $x=-3$

When $x=-3$, the denominator $(x + 2)(x + 3)=(-3 + 2)(-3+3)=0$ and the numerator is non - zero. So $x=-3$ is an asymptote.

Step4: Analyze $x=-2$

When $x=-2$, the denominator $(x + 2)(x + 3)=(-2 + 2)(-2+3)=0$ and the numerator is non - zero. So $x=-2$ is an asymptote.

Step5: Analyze $x = 0$

Since we canceled out the factor $x$ in the simplification process, $x = 0$ is a hole.

Step6: Analyze $x=2,3,5$

When $x = 2,3,5$, the denominator $(x + 2)(x + 3)\neq0$. So $x = 2,3,5$ are neither asymptotes nor holes.

Answer:

$x=-3$: asymptote $x=-2$: asymptote $x = 0$: hole $x = 2$: neither $x = 3$: neither $x = 5$: neither