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. $x=-9$ (use comma to separate answers as needed)
b. there is no local maximum

Explanation:

Step1: Find critical points

Critical points occur where \(f^{\prime}(x) = 0\) or \(f^{\prime}(x)\) is undefined.
Set \(f^{\prime}(x)=\frac{(x - 5)(x + 9)}{(x + 3)(x - 7)}=0\). Then \((x - 5)(x + 9)=0\), so \(x=5\) or \(x=-9\). \(f^{\prime}(x)\) is undefined at \(x=-3\) and \(x = 7\) (but these are not in the domain of \(f^{\prime}(x)\) for critical - point consideration as per the problem's \(x
eq-3,7\) condition). So critical points are \(x=-9\) and \(x = 5\).

Step2: Determine intervals of increase and decrease

Use a sign - chart. The critical points \(x=-9\) and \(x = 5\) divide the real line (excluding \(x=-3\) and \(x = 7\)) into 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 \(f^{\prime}(0)=\frac{(0 - 5)(0 + 9)}{(0+3)(0 - 7)}=\frac{(-5)(9)}{(3)(-7)}=\frac{-45}{-21}>0\).
• For \(x\in(5,7)\), let \(x = 6\). Then \(f^{\prime}(6)=\frac{(6 - 5)(6 + 9)}{(6+3)(6 - 7)}=\frac{(1)(15)}{(9)(-1)}=\frac{15}{-9}<0\).
• For \(x\in(7,\infty)\), let \(x = 8\). Then \(f^{\prime}(8)=\frac{(8 - 5)(8 + 9)}{(8+3)(8 - 7)}=\frac{(3)(17)}{(11)(1)}=\frac{51}{11}>0\).

So \(f(x)\) is increasing on \((-\infty,-9)\cup(-3,5)\cup(7,\infty)\) and decreasing on \((-9,-3)\cup(5,7)\).

Step3: Find local maxima and minima

By the first - derivative test:

• At \(x=-9\), \(f^{\prime}(x)\) changes from positive (on \((-\infty,-9)\)) to negative (on \((-9,-3)\)), so \(x=-9\) is a local maximum.
• At \(x = 5\), \(f^{\prime}(x)\) changes from positive (on \((-3,5)\)) to negative (on \((5,7)\)), so \(x = 5\) is a local minimum.

Answer:

a. The critical points of \(f\) are \(x=-9\) and \(x = 5\).
b. The function \(f\) is increasing on the intervals \((-\infty,-9)\cup(-3,5)\cup(7,\infty)\) and decreasing on \((-9,-3)\cup(5,7)\).
c. The function \(f\) has a local maximum at \(x=-9\) and a local minimum at \(x = 5\).