Question

let f be the function defined by $f(x)=\frac{3x^{3}+2x^{2}}{x^{2}-x}$. which of the following statements is true?
a f has a discontinuity due to a vertical asymptote at $x = 0$ and at $x = 1$.
b f has a removable discontinuity at $x = 0$ and a jump discontinuity at $x = 1$.
c f has a removable discontinuity at $x = 0$ and a discontinuity due to a vertical asymptote at $x = 1$.
d f is continuous at $x = 0$, and f has a discontinuity due to a vertical asymptote at $x = 1$.

Explanation:

Step1: Simplify the function

First, factor the numerator and denominator. Given $f(x)=\frac{3x^{3}+2x^{2}}{x^{2}-x}=\frac{x^{2}(3x + 2)}{x(x - 1)}$. For $x
eq0$, we can simplify it to $f(x)=\frac{x(3x + 2)}{x - 1}=\frac{3x^{2}+2x}{x - 1}$.

Step2: Analyze the discontinuity at $x = 0$

The original function $f(x)$ is undefined at $x = 0$. But $\lim_{x
ightarrow0}\frac{3x^{3}+2x^{2}}{x^{2}-x}=\lim_{x
ightarrow0}\frac{x(3x + 2)}{x - 1}=0$. Since the limit exists as $x
ightarrow0$, $x = 0$ is a removable - discontinuity.

Step3: Analyze the discontinuity at $x = 1$

$\lim_{x
ightarrow1^{+}}\frac{3x^{3}+2x^{2}}{x^{2}-x}=\lim_{x
ightarrow1^{+}}\frac{x(3x + 2)}{x - 1}=+\infty$ and $\lim_{x
ightarrow1^{-}}\frac{3x^{3}+2x^{2}}{x^{2}-x}=\lim_{x
ightarrow1^{-}}\frac{x(3x + 2)}{x - 1}=-\infty$. Since the limit does not exist and goes to infinity as $x
ightarrow1$, $x = 1$ is a vertical - asymptote.

Answer:

C. $f$ has a removable discontinuity at $x = 0$ and a discontinuity due to a vertical asymptote at $x = 1$.