what are the open intervals of increase and decrease for the rational function f(x)=10/(x^2 - 3x - 18)?

Explanation:

Step1: Factor the denominator

First, factor $x^{2}-3x - 18=(x - 6)(x+3)$. The function $f(x)=\frac{10}{(x - 6)(x + 3)}$ is undefined at $x=-3$ and $x = 6$.

Step2: Find the derivative

Use the quotient - rule. If $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 10$, $u^\prime=0$, $v=(x - 6)(x + 3)=x^{2}-3x - 18$, $v^\prime=2x-3$. So $f^\prime(x)=\frac{0\times(x^{2}-3x - 18)-10\times(2x - 3)}{(x^{2}-3x - 18)^{2}}=\frac{-10(2x - 3)}{(x^{2}-3x - 18)^{2}}=\frac{-20x + 30}{(x - 6)^{2}(x + 3)^{2}}$.

Step3: Find critical points

Set $f^\prime(x)=0$. Then $-20x+30 = 0$, which gives $x=\frac{3}{2}$.

Step4: Test intervals

We have the intervals $(-\infty,-3)$, $(-3,\frac{3}{2})$, $(\frac{3}{2},6)$ and $(6,\infty)$. For $x\in(-\infty,-3)$, let $x=-4$. Then $f^\prime(-4)=\frac{-20\times(-4)+30}{((-4 - 6)^{2}(-4 + 3)^{2})}=\frac{80 + 30}{100\times1}>0$. For $x\in(-3,\frac{3}{2})$, let $x = 0$. Then $f^\prime(0)=\frac{-20\times0+30}{(0 - 6)^{2}(0 + 3)^{2}}=\frac{30}{36\times9}>0$. For $x\in(\frac{3}{2},6)$, let $x = 3$. Then $f^\prime(3)=\frac{-20\times3+30}{(3 - 6)^{2}(3 + 3)^{2}}=\frac{-60 + 30}{9\times36}<0$. For $x\in(6,\infty)$, let $x = 7$. Then $f^\prime(7)=\frac{-20\times7+30}{(7 - 6)^{2}(7 + 3)^{2}}=\frac{-140+30}{1\times100}<0$.

Answer:

The function $f(x)$ is increasing on the intervals $(-\infty,-3)$ and $(-3,\frac{3}{2})$, and decreasing on the intervals $(\frac{3}{2},6)$ and $(6,\infty)$.