what are the open intervals of decrease for the given rational function below\n\n$f(x)=\frac{x}{x^{3}-6x^{2}+…

Answer:

A. $(3, 5)\cup(5,\infty)$

Explanation:

Step1: Simplify the function

First, factor the denominator: $x^{3}-6x^{2}+5x=x(x^{2}-6x + 5)=x(x - 1)(x - 5)$. So $f(x)=\frac{x}{x(x - 1)(x - 5)}=\frac{1}{(x - 1)(x - 5)}=\frac{1}{x^{2}-6x + 5}$ for $x\neq0$.

Step2: Find the derivative

Using the quotient - rule or rewrite as $(x^{2}-6x + 5)^{-1}$ and use the chain - rule. If $y=(x^{2}-6x + 5)^{-1}$, then $y^\prime=-\frac{2x - 6}{(x^{2}-6x + 5)^{2}}$.

Step3: Find critical points

Set $y^\prime = 0$. Then $2x-6 = 0$, so $x = 3$. Also, the function is undefined at $x=1$ and $x = 5$.

Step4: Test intervals

Test the intervals $(-\infty,1)$, $(1,3)$, $(3,5)$ and $(5,\infty)$ using a test - point in each interval. For example, for the interval $(3,5)$, let $x = 4$, then $y^\prime=-\frac{2\times4 - 6}{(4^{2}-6\times4 + 5)^{2}}=-\frac{2}{1}<0$, so the function is decreasing on $(3,5)$. For the interval $(5,\infty)$, let $x=6$, then $y^\prime=-\frac{2\times6 - 6}{(6^{2}-6\times6 + 5)^{2}}=-\frac{6}{25}<0$, so the function is decreasing on $(5,\infty)$.