Question

consider the graph of the function $f(x)=\frac{x^{3}-x^{2}-2x}{x^{2}+x}$. which is a removable discontinuity for the graph? select all that apply. select all that apply: $x = - 2$ $x = - 1$ $x = 0$ $x = 1$ $x = 2$

Explanation:

Step1: Factor the function

First, factor the numerator and denominator.
The numerator $x^{3}-x^{2}-2x=x(x - 2)(x + 1)$ and the denominator $x^{2}+x=x(x + 1)$. So, $f(x)=\frac{x(x - 2)(x + 1)}{x(x + 1)}$, $x
eq0,x
eq - 1$.

Step2: Simplify the function

Cancel out the common factors $x$ and $(x + 1)$ (for $x
eq0$ and $x
eq - 1$). We get $f(x)=x - 2$ for $x
eq0,x
eq - 1$.
A removable discontinuity occurs when a factor in the denominator can be canceled out with a factor in the numerator.
The values of $x$ that make the original denominator zero and can be canceled are $x = 0$ and $x=-1$.

Answer:

$x=-1$, $x = 0$