Question

1. find and classify (jump, removable or infinite) the discontinuities for the function

g(x)=\frac{x^{2}-5x + 6}{x^{3}-9x}

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{2}-5x + 6=(x - 2)(x - 3)$. The denominator $x^{3}-9x=x(x^{2}-9)=x(x - 3)(x+3)$. So $g(x)=\frac{(x - 2)(x - 3)}{x(x - 3)(x + 3)}$.

Step2: Find the points of discontinuity

The function $g(x)$ is discontinuous when the denominator is zero. Set $x(x - 3)(x + 3)=0$. The solutions are $x = 0,x=3,x=-3$.

Step3: Simplify the function

Cancel out the common factor $(x - 3)$ (for $x
eq3$), we get $g(x)=\frac{x - 2}{x(x + 3)},x
eq3$.

Step4: Classify the discontinuities

• For $x = 3$:

Since $\lim_{x
ightarrow3}g(x)=\lim_{x
ightarrow3}\frac{x - 2}{x(x + 3)}=\frac{3-2}{3\times(3 + 3)}=\frac{1}{18}$, the discontinuity at $x = 3$ is removable.

• For $x=0$:

$\lim_{x
ightarrow0^{+}}g(x)=\lim_{x
ightarrow0^{+}}\frac{x - 2}{x(x + 3)}=-\infty$ and $\lim_{x
ightarrow0^{-}}g(x)=\lim_{x
ightarrow0^{-}}\frac{x - 2}{x(x + 3)}=\infty$. So the discontinuity at $x = 0$ is infinite.

• For $x=-3$:

$\lim_{x
ightarrow - 3^{+}}g(x)=\lim_{x
ightarrow - 3^{+}}\frac{x - 2}{x(x + 3)}=-\infty$ and $\lim_{x
ightarrow - 3^{-}}g(x)=\lim_{x
ightarrow - 3^{-}}\frac{x - 2}{x(x + 3)}=\infty$. So the discontinuity at $x=-3$ is infinite.

Answer:

The function $g(x)$ has a removable discontinuity at $x = 3$ and infinite discontinuities at $x=0$ and $x=-3$.