Question

due wednesday by 11:59pm points 100 submitting an external tool consider the graph of the function $f(x)=\frac{x^{3}-6x^{2}+8x}{x^{2}-2x - 8}$. which is a removable discontinuity for the graph? select all that apply. select all that apply: $x=-4$ $x=-2$ $x = 0$ $x = 2$ $x = 4$

Explanation:

Step1: Factor the numerator and denominator

The numerator $x^{3}-6x^{2}+8x=x(x - 2)(x - 4)$. The denominator $x^{2}-2x - 8=(x - 4)(x+2)$. So $f(x)=\frac{x(x - 2)(x - 4)}{(x - 4)(x + 2)}$ for $x
eq4,x
eq - 2$.

Step2: Identify removable discontinuities

A removable discontinuity occurs when a factor in the denominator can be canceled out in the numerator. After canceling out the common factor $(x - 4)$ (when $x
eq4$), the function simplifies. The values of $x$ for which the original function is undefined but the simplified - function is defined are the removable discontinuities. The original function is undefined at $x=-2,x = 4$. But since we can cancel the $(x - 4)$ factor, $x = 4$ is a removable discontinuity. Also, when we consider the fact that the function is a rational function and we analyze the behavior around other points where the numerator and denominator have zeros, we find that $x=-2,x = 0,x = 2$ are not points that make the function have a non - removable discontinuity in the sense of vertical asymptotes. They are removable discontinuities as we can re - define the function at these points to make it continuous.

Answer:

$x=-4$ is not a removable discontinuity. $x=-2$ is a removable discontinuity. $x = 0$ is a removable discontinuity. $x = 2$ is a removable discontinuity. $x = 4$ is a removable discontinuity. So the correct answers are: $x=-2$, $x = 0$, $x = 2$, $x = 4$.