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, if necessary, fill in the answer box within your choice.\na. $x=-9$ (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, if necessary, fill in the answer box within your choice.\na. $x=-9$ (use comma to separate answers as needed)\nb. there is no local maximum

Answer

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\neq-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).