question 39 (1 point) find the x - value (if any) at which f(x) is not continuous. f(x)=(x - 4)/(x^2 - 2x - 8)

Explanation:

Step1: Factor the denominator

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

Step2: Determine the non - continuous points

A rational function $y=\frac{g(x)}{h(x)}$ is discontinuous where $h(x)=0$. Set $(x - 4)(x + 2)=0$. Solving the equation $(x - 4)(x + 2)=0$ using the zero - product property (if $ab = 0$, then $a=0$ or $b = 0$), we get $x=4$ or $x=-2$. But when $x = 4$, we can simplify $f(x)$ by canceling out the common factor $(x - 4)$ (for $x\neq4$), and $f(x)=\frac{1}{x + 2}$ for $x\neq4$. The limit as $x\rightarrow4$ exists. However, when $x=-2$, the denominator is zero and the numerator is non - zero.

Answer:

$x=-2$