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 increasing on the interval(s) $(-\\infty,-9),(-3,5),(7,\\infty)$\n(type your answer in interval notation. use a comma to separate answers as needed )\nb. the function f is not increasing anywhere.\non what open intervals is f decreasing? select the correct choice below and, if necessary, fill in the answer box within your choice.\na. the function f is decreasing on the interval(s)

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 increasing on the interval(s) $(-\\infty,-9),(-3,5),(7,\\infty)$\n(type your answer in interval notation. use a comma to separate answers as needed )\nb. the function f is not increasing anywhere.\non what open intervals is f decreasing? select the correct choice below and, if necessary, fill in the answer box within your choice.\na. the function f is decreasing on the interval(s)

Answer

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))