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 decreasing on the interval(s) $(-9,-3),(5,7)$
(type your answer in interval notation. use a comma to separate answers as needed.)
b. the function $f$ is not decreasing anywhere.
c. at what points, if any, does $f$ assume local maximum values? select the correct choice below and,
if necessary, fill in the answer box within your choice.
a. $x=$ (use comma to separate answers as needed)
b. there is no local maximum.

Explanation:

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

We have \(f^{\prime}(x)=\frac{(x - 5)(x + 9)}{(x + 3)(x - 7)}\), \(x
eq-3,7\). The critical points are \(x=-9,x = 5\) (where \(f^{\prime}(x)=0\)).
We use a sign - chart. Consider the intervals \((-\infty,-9),(-9,-3),(-3,5),(5,7),(7,\infty)\).
For \(x\in(-\infty,-9)\), 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 \(x\in(-9,-3)\), 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 \(x\in(-3,5)\), 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 \(x\in(5,7)\), 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 \(x\in(7,\infty)\), 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 local maxima

By the first - derivative test, if \(f^{\prime}(x)\) changes sign from positive to negative at a critical point \(x = c\), then \(f(x)\) has a local maximum at \(x = c\).
Since \(f^{\prime}(x)\) changes sign from positive to negative at \(x=-9\) (from the interval \((-\infty,-9)\) to \((-9,-3)\)), \(f(x)\) has a local maximum at \(x=-9\).

Answer:

A. \(x=-9\)