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) $(-\\infty,-9),(-9,5)$ (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) $(-\\infty,-9),(-9,5)$ (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: 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\neq-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.