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. $x=-9,5$ (use comma to separate answers as needed)
b. the function $f$ has no critical points.
b. on what open intervals is $f$ increasing? select the correct choice below and, if necessary, fill in the answer box within your choice.
a. the function $f$ is increasing on the interval(s) $(-\infty,-9),(-9,5)$ (type your answer in interval notation. use a comma to separate answers as needed)
b. the function $f$ is not increasing anywhere.

Explanation:

Step1: Determine critical points

Critical points occur where \(f^{\prime}(x) = 0\) or \(f^{\prime}(x)\) is undefined. Set \(f^{\prime}(x)=\frac{(x - 5)(x + 9)}{(x + 3)(x - 7)}=0\). Numerator \((x - 5)(x + 9)=0\) gives \(x = 5\) or \(x=-9\). \(x=-3\) and \(x = 7\) make \(f^{\prime}(x)\) undefined but are not in the domain of \(f(x)\) (since \(x
eq-3,7\)). So critical points are \(x=-9,5\).

Step2: Test intervals for increasing/decreasing

Use test - points in intervals \((-\infty,-9)\), \((-9,-3)\), \((-3,5)\), \((5,7)\), \((7,\infty)\).

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

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

Step3: Determine local maxima/minima

By First - Derivative Test:

• At \(x=-9\): \(f^{\prime}(x)\) changes from positive (left of \(x = - 9\)) to negative (right of \(x=-9\)), so \(x=-9\) is a local maximum.
• At \(x = 5\): \(f^{\prime}(x)\) changes from positive (left of \(x = 5\)) to negative (right of \(x = 5\)), so \(x = 5\) is a local maximum.

Answer:

a. \(x=-9,5\)
b. The function \(f\) is increasing on the intervals \((-\infty,-9),(-3,5),(7,\infty)\)
c. Local maxima at \(x=-9\) and \(x = 5\), no local minima.