consider the graph of the function $f(x)=\frac{x^{3}-6x^{2}+8x}{x^{2}-2x - 8}$. which is a removable…

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: $x = - 4$ $x=-2$ $x = 0$ $x = 2$ $x = 4$

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: $x = - 4$ $x=-2$ $x = 0$ $x = 2$ $x = 4$

Answer

Answer:

x = 2

Explanation:

Step1: Factor the numerator and denominator

Factor $x^{3}-6x^{2}+8x=x(x - 2)(x - 4)$ and $x^{2}-2x - 8=(x - 4)(x+2)$. So $f(x)=\frac{x(x - 2)(x - 4)}{(x - 4)(x + 2)}$.

Step2: Identify removable discontinuities

A removable discontinuity occurs when a factor in the numerator and denominator cancels out. Here, the factor $(x - 4)$ cancels out for $x\neq4$. After canceling, the function is undefined at $x=-2$ (vertical - asymptote) and has a removable discontinuity at $x = 2$ since the original function is not defined at $x = 2$ but the limit as $x\rightarrow2$ exists. The function is well - behaved at $x=0$ and $x=-4$ is not a point of discontinuity as it does not make the denominator zero.