(4) 11. find and classify the discontinuities for the function\ng(x)=\begin{cases}\frac{x^{2}-3x}{x^{3}-2x^{2…

Explanation:

Step1: Simplify the function for $x\neq0$

First, factor the numerator and denominator of $\frac{x^{2}-3x}{x^{3}-2x^{2}-3x}$. Numerator: $x^{2}-3x=x(x - 3)$. Denominator: $x^{3}-2x^{2}-3x=x(x^{2}-2x - 3)=x(x - 3)(x+1)$. So, for $x\neq0$, $g(x)=\frac{x(x - 3)}{x(x - 3)(x + 1)}=\frac{1}{x + 1}$.

Step2: Check the limit as $x\rightarrow0$

Find $\lim_{x\rightarrow0}g(x)$. Since for $x\neq0$, $g(x)=\frac{1}{x + 1}$, then $\lim_{x\rightarrow0}g(x)=\lim_{x\rightarrow0}\frac{1}{x + 1}=1$. And $g(0)=1$. So, there is no discontinuity at $x = 0$.

Step3: Check for other discontinuities

Set the denominator of the simplified - function $y=\frac{1}{x + 1}$ equal to zero. $x+1 = 0$ gives $x=-1$. $\lim_{x\rightarrow - 1^{-}}\frac{1}{x + 1}=-\infty$ and $\lim_{x\rightarrow - 1^{+}}\frac{1}{x + 1}=\infty$.

Answer:

a) None b) None c) $x=-1$