Question

answer the questions below about the function whose derivative is $f(x)=\frac{(x - 2)(x + 8)}{(x + 1)(x - 3)},x\
eq-1,3$.
a. what are the critical points of f?
b. on what open intervals is f increasing or decreasing?
c. at what points, if any, does f assume local maximum and minimum values?
a. $x=-8,2$ (use comma to separate answers as needed)
b. there is no local maximum.
at what points, if any, does f assume local minimum values? select the correct choice below and,
if necessary, fill in the answer box within your choice.
a. $x=\square$ (use comma to separate answers as needed)
b. there is no local minimum.

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.