Question

answer the questions below about the function whose derivative is $f(x)=\frac{(x - 2)(x + 8)}{(x + 1)(x - 3)},x\
eq-1,3$.

a. what are the critical points of f?
b. on what open intervals is f increasing or decreasing?
c. at what points, if any, does f assume local maximum and minimum values?

a. the function f is increasing on the interval(s) $(-\infty,-8),(-1,2),(3,\infty)$
(type your answer in interval notation. use a comma to separate answers as needed )
b. the function f is not increasing anywhere.
on what open intervals is f decreasing? select the correct choice below and, if necessary, fill in the answer box within your choice.
a. the function f is decreasing on the interval(s)
(type your answer in interval notation. use a comma to separate answers as needed )
b. the function f is not decreasing anywhere.

Explanation:

Step1: Analyze the sign of \(f^{\prime}(x)\)

We have \(f^{\prime}(x)=\frac{(x - 2)(x + 8)}{(x + 1)(x - 3)}\). The critical points (where \(f^{\prime}(x)=0\) or is undefined) are \(x=-8,x=-1,x = 2,x=3\). These points divide the real - line into intervals \((-\infty,-8),(-8,-1),(-1,2),(2,3),(3,\infty)\).
Take test points:

• For the interval \((-\infty,-8)\), let \(x=-9\). Then \(f^{\prime}(-9)=\frac{(-9 - 2)(-9 + 8)}{(-9 + 1)(-9 - 3)}=\frac{(-11)(-1)}{(-8)(-12)}=\frac{11}{96}>0\).
• For the interval \((-8,-1)\), let \(x=-2\). Then \(f^{\prime}(-2)=\frac{(-2 - 2)(-2 + 8)}{(-2 + 1)(-2 - 3)}=\frac{(-4)(6)}{(-1)(-5)}=\frac{-24}{5}<0\).
• For the interval \((-1,2)\), let \(x = 0\). Then \(f^{\prime}(0)=\frac{(0 - 2)(0 + 8)}{(0 + 1)(0 - 3)}=\frac{(-2)(8)}{(1)(-3)}=\frac{16}{3}>0\).
• For the interval \((2,3)\), let \(x=\frac{5}{2}\). Then \(f^{\prime}(\frac{5}{2})=\frac{(\frac{5}{2}-2)(\frac{5}{2}+8)}{(\frac{5}{2}+1)(\frac{5}{2}-3)}=\frac{(\frac{1}{2})(\frac{21}{2})}{(\frac{7}{2})(-\frac{1}{2})}=-\frac{21}{7}=-3<0\).
• For the interval \((3,\infty)\), let \(x = 4\). Then \(f^{\prime}(4)=\frac{(4 - 2)(4 + 8)}{(4 + 1)(4 - 3)}=\frac{(2)(12)}{(5)(1)}=\frac{24}{5}>0\).

Step2: Determine the decreasing intervals

Since \(f^{\prime}(x)<0\) on the intervals \((-8,-1)\) and \((2,3)\).

Answer:

A. The function \(f\) is decreasing on the interval(s) \((-8,-1),(2,3)\)