what are the open intervals of increase for the given rational function below? f(x)=x(x + 2)/x(x² + x - 2) o…

Explanation:

Step1: Simplify the function

First, factor the denominator $x(x^{2}+x - 2)=x(x + 2)(x - 1)$. The function $f(x)=\frac{x(x + 2)}{x(x^{2}+x - 2)}=\frac{x(x + 2)}{x(x + 2)(x - 1)}$. After canceling out $x(x + 2)$ (for $x\neq0$ and $x\neq - 2$), we get $f(x)=\frac{1}{x - 1},x\neq0,x\neq - 2$.

Step2: Find the derivative

Using the quotient - rule, if $y=\frac{1}{x - 1}=(x - 1)^{-1}$, then $y^\prime=-1\times(x - 1)^{-2}=-\frac{1}{(x - 1)^{2}}$.

Step3: Determine intervals of increase

A function $y = f(x)$ is increasing when $y^\prime>0$. But $y^\prime=-\frac{1}{(x - 1)^{2}}<0$ for all $x\neq1$. So no intervals of increase exist for this function.

Answer:

No intervals of increase exist for this function.