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

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

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). Using the zero - product property (a\times b = 0) implies (a = 0) or (b=0), so (x-5 = 0) gives (x = 5) and (x + 9=0) gives (x=-9). The derivative is undefined at (x=-3) and (x = 7), but these are not in the domain of the original function (since the derivative is a rational function and the original function's domain excludes (x=-3) and (x = 7)). So the critical points are (x=-9) and (x = 5).

Step2: Test intervals for increasing/decreasing

We use the critical points (x=-9) and (x = 5) to divide the number line into intervals: ((-\infty,-9)), ((-9,5)), ((5,\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 ((-9,5)), 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,\infty)), 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).

Answer:

a. (x=-9,5) b. The function (f) is increasing on the intervals ((-\infty,-9)) and ((-9,5))