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

answer the questions below about the function whose derivative is $f(x)=\\frac{(x - 5)(x + 9)}{(x + 3)(x - 7)},x\\neq-3,7$.\na. what are the critical points of $f$?\nb. on what open intervals is $f$ increasing or decreasing?\nc. at what points, if any, does $f$ assume local maximum and minimum values?\na. the function $f$ is decreasing on the interval(s) $(-9,-3),(5,7)$\n(type your answer in interval notation. use a comma to separate answers as needed.)\nb. the function $f$ is not decreasing anywhere.\nc. at what points, if any, does $f$ assume local maximum values? select the correct choice below and,\nif necessary, fill in the answer box within your choice.\na. $x=$ (use comma to separate answers as needed)\nb. there is no local maximum.

answer the questions below about the function whose derivative is $f(x)=\\frac{(x - 5)(x + 9)}{(x + 3)(x - 7)},x\\neq-3,7$.\na. what are the critical points of $f$?\nb. on what open intervals is $f$ increasing or decreasing?\nc. at what points, if any, does $f$ assume local maximum and minimum values?\na. the function $f$ is decreasing on the interval(s) $(-9,-3),(5,7)$\n(type your answer in interval notation. use a comma to separate answers as needed.)\nb. the function $f$ is not decreasing anywhere.\nc. at what points, if any, does $f$ assume local maximum values? select the correct choice below and,\nif necessary, fill in the answer box within your choice.\na. $x=$ (use comma to separate answers as needed)\nb. there is no local maximum.

Answer

Explanation:

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

We have (f^{\prime}(x)=\frac{(x - 5)(x + 9)}{(x + 3)(x - 7)}), (x\neq-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)