Question

answer the questions below about the function whose derivative is $f(x)=\frac{(x - 5)(x + 9)}{(x + 3)(x - 7)},x\
eq-3,7$.
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) $(-\infty,-9),(-3,5),(7,\infty)$
(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)

Explanation:

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

We have \(f^{\prime}(x)=\frac{(x - 5)(x + 9)}{(x + 3)(x - 7)}\). The critical points (where \(f^{\prime}(x)=0\) or \(f^{\prime}(x)\) is undefined) are \(x=-9,x = 5,x=-3,x = 7\).
We use the test - point method. Consider the intervals \((-\infty,-9),(-9,-3),(-3,5),(5,7),(7,\infty)\)

• For the interval \((-\infty,-9)\), let \(x=-10\). Then \(f^{\prime}(-10)=\frac{(-10 - 5)(-10 + 9)}{(-10+3)(-10 - 7)}=\frac{(-15)(-1)}{(-7)(-17)}=\frac{15}{119}>0\)
• For the interval \((-9,-3)\), let \(x=-4\). Then \(f^{\prime}(-4)=\frac{(-4 - 5)(-4 + 9)}{(-4+3)(-4 - 7)}=\frac{(-9)(5)}{(-1)(-11)}=\frac{-45}{11}<0\)
• For the interval \((-3,5)\), let \(x = 0\). Then \(f^{\prime}(0)=\frac{(0 - 5)(0 + 9)}{(0+3)(0 - 7)}=\frac{(-5)(9)}{(3)(-7)}=\frac{-45}{-21}=\frac{15}{7}>0\)
• For the interval \((5,7)\), let \(x=6\). Then \(f^{\prime}(6)=\frac{(6 - 5)(6 + 9)}{(6+3)(6 - 7)}=\frac{(1)(15)}{(9)(-1)}=-\frac{5}{3}<0\)
• For the interval \((7,\infty)\), let \(x = 8\). Then \(f^{\prime}(8)=\frac{(8 - 5)(8 + 9)}{(8+3)(8 - 7)}=\frac{(3)(17)}{(11)(1)}=\frac{51}{11}>0\)

Step2: Determine the decreasing intervals

Since \(f^{\prime}(x)<0\) on the intervals \((-9,-3)\) and \((5,7)\)

Answer:

The function \(f\) is decreasing on the interval(s) \((-9,-3),(5,7)\)