answer the questions below about the function whose derivative is $f(x)=\\frac{(x - 2)(x + 8)}{(x + 1)(x…

answer the questions below about the function whose derivative is $f(x)=\\frac{(x - 2)(x + 8)}{(x + 1)(x - 3)},x\\neq-1,3$.\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. $x=-8,2$ (use comma to separate answers as needed)\nb. there is no local maximum.\nat what points, if any, does f assume local minimum values? select the correct choice below and,\nif necessary, fill in the answer box within your choice.\na. $x=\\square$ (use comma to separate answers as needed)\nb. there is no local minimum.

answer the questions below about the function whose derivative is $f(x)=\\frac{(x - 2)(x + 8)}{(x + 1)(x - 3)},x\\neq-1,3$.\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. $x=-8,2$ (use comma to separate answers as needed)\nb. there is no local maximum.\nat what points, if any, does f assume local minimum values? select the correct choice below and,\nif necessary, fill in the answer box within your choice.\na. $x=\\square$ (use comma to separate answers as needed)\nb. there is no local minimum.

Answer

Explanation:

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

We have (f^{\prime}(x)=\frac{(x - 2)(x + 8)}{(x + 1)(x - 3)}). The critical points (where (f^{\prime}(x)=0)) are (x=-8) and (x = 2) (from the numerator ((x - 2)(x + 8)=0)). The function (f^{\prime}(x)) is undefined at (x=-1) and (x = 3). We consider the intervals ((-\infty,-8)), ((-8,-1)), ((-1,2)), ((2,3)) and ((3,\infty)).

  • For (x\in(-\infty,-8)), let (x=-9), then (f^{\prime}(-9)=\frac{(-9 - 2)(-9+8)}{(-9 + 1)(-9 - 3)}=\frac{(-11)(-1)}{(-8)(-12)}=\frac{11}{96}>0).
  • For (x\in(-8,-1)), let (x=-2), then (f^{\prime}(-2)=\frac{(-2 - 2)(-2 + 8)}{(-2+1)(-2 - 3)}=\frac{(-4)(6)}{(-1)(-5)}=-\frac{24}{5}<0).
  • For (x\in(-1,2)), let (x=0), then (f^{\prime}(0)=\frac{(0 - 2)(0 + 8)}{(0 + 1)(0 - 3)}=\frac{(-2)(8)}{(1)(-3)}=\frac{16}{3}>0).
  • For (x\in(2,3)), let (x=\frac{5}{2}), then (f^{\prime}(\frac{5}{2})=\frac{(\frac{5}{2}-2)(\frac{5}{2}+8)}{(\frac{5}{2}+1)(\frac{5}{2}-3)}=\frac{(\frac{1}{2})(\frac{21}{2})}{(\frac{7}{2})(-\frac{1}{2})}=-\frac{21}{7}=-3<0).
  • For (x\in(3,\infty)), let (x = 4), then (f^{\prime}(4)=\frac{(4 - 2)(4 + 8)}{(4 + 1)(4 - 3)}=\frac{(2)(12)}{(5)(1)}=\frac{24}{5}>0).

Step2: Determine local minima

By the first - derivative test: A function (y = f(x)) has a local minimum at a critical point (x = c) if (f^{\prime}(x)) changes sign from negative to positive at (x = c). We see that (f^{\prime}(x)) changes sign from negative to positive at (x=-8) (from (x\in(-8,-1)) ((f^{\prime}(x)<0)) to (x\in(-\infty,-8)) ((f^{\prime}(x)>0)) is wrong, actually from (x\in(-\infty,-8)) ((f^{\prime}(x)>0)) to (x\in(-8,-1)) ((f^{\prime}(x)<0)) is wrong. Wait, no: We check the sign change:

  • At (x=-8): (f^{\prime}(x)) changes from positive (when (x\in(-\infty,-8))) to negative (when (x\in(-8,-1))), so (x=-8) is a local maximum.
  • At (x = 2): (f^{\prime}(x)) changes from positive (when (x\in(-1,2))) to negative (when (x\in(2,3))), so (x = 2) is a local maximum.

Answer:

B. There is no local minimum.