what are the open intervals of decrease for the given rational function below? f(x)=(x - 2)/(x^2 - 2x)…

Explanation:

Step1: Simplify the function

First, factor the denominator: $f(x)=\frac{x - 2}{x(x - 2)}=\frac{1}{x},x\neq0,2$.

Step2: Find the derivative

The derivative of $y=\frac{1}{x}=x^{-1}$ using the power - rule $(x^n)'=nx^{n - 1}$ is $y'=-x^{-2}=-\frac{1}{x^{2}}$.

Step3: Determine where the function is decreasing

A function $y = f(x)$ is decreasing when $y'<0$. Since $y'=-\frac{1}{x^{2}}<0$ for all $x\neq0$. The domain of the original function (after simplification and considering the original denominator) is $(-\infty,0)\cup(0,2)\cup(2,\infty)$. And the function $y = f(x)$ is decreasing on $(-\infty,0)\cup(0,2)\cup(2,\infty)$.

Answer:

$(-\infty,0)\cup(0,2)\cup(2,\infty)$