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?\n\na. the function f is decreasing on the interval(s) $(-8,-1),(2,3)$.\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. $x=$ (use comma to separate answers as needed)\nb. there is no local maximum.

answer the questions below about the function whose derivative is $f(x)=\\frac{(x - 2)(x + 8)}{(x + 1)(x - 3)},x\\neq-1,3$.\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?\n\na. the function f is decreasing on the interval(s) $(-8,-1),(2,3)$.\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. $x=$ (use comma to separate answers as needed)\nb. there is no local maximum.

Answer

Explanation:

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

We know that (f^{\prime}(x)=\frac{(x - 2)(x + 8)}{(x + 1)(x - 3)}). The critical points (where (f^{\prime}(x)=0) or (f^{\prime}(x)) is undefined) are (x=-8,x=-1,x = 2,x=3). We use a sign - chart. Consider the intervals ((-\infty,-8),(-8,-1),(-1,2),(2,3),(3,\infty)) Take a test point in each interval:

  • For (x=-9) (in ((-\infty,-8))): (f^{\prime}(-9)=\frac{(-9 - 2)(-9+8)}{(-9 + 1)(-9 - 3)}=\frac{(-11)(-1)}{(-8)(-12)}=\frac{11}{96}>0)
  • For (x =-\frac{1}{2}) (in ((-8,-1))): (f^{\prime}(-\frac{1}{2})=\frac{(-\frac{1}{2}-2)(-\frac{1}{2}+8)}{(-\frac{1}{2}+1)(-\frac{1}{2}-3)}=\frac{(-\frac{5}{2})(\frac{15}{2})}{(\frac{1}{2})(-\frac{7}{2})}=\frac{-\frac{75}{4}}{-\frac{7}{4}}=\frac{75}{7}>0) (Wait, no! Let's recalculate: (f^{\prime}(x)=\frac{(x - 2)(x + 8)}{(x + 1)(x - 3)}). For (x=-4.5) (in ((-8,-1))): (f^{\prime}(-4.5)=\frac{(-4.5-2)(-4.5 + 8)}{(-4.5+1)(-4.5-3)}=\frac{(-6.5)(3.5)}{(-3.5)(-7.5)}=\frac{-22.75}{26.25}<0)
  • For (x=0) (in ((-1,2))): (f^{\prime}(0)=\frac{(0 - 2)(0 + 8)}{(0 + 1)(0 - 3)}=\frac{(-2)(8)}{(1)(-3)}=\frac{-16}{-3}>0)
  • For (x=\frac{5}{2}) (in ((2,3))): (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{\frac{21}{4}}{-\frac{7}{4}}=- 3<0)
  • For (x = 4) (in ((3,\infty))): (f^{\prime}(4)=\frac{(4 - 2)(4 + 8)}{(4 + 1)(4 - 3)}=\frac{(2)(12)}{(5)(1)}=\frac{24}{5}>0)

A function (y = f(x)) has a local maximum when (f^{\prime}(x)) changes from positive to negative. Since (f^{\prime}(x)) changes from positive to negative at (x = 2)

Answer:

A. (x = 2)