Question

answer the following questions about the function whose derivative is $f(x)=\frac{x^{2}(x - 2)}{x + 5},x\
eq-5$.
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 or minimum values?
b. on what open intervals is f increasing or decreasing? select the correct choice below and, if necessary, fill in the answer box(es) to complete your choice.
○ a. the function f is increasing on the open interval(s) and never decreasing
(type your answer in interval notation. use a comma to separate answers as needed.)
○ b. the function f is increasing on the open interval(s) $(2,\infty)$ and decreasing on the open interval(s)
$(-\infty,2)$
(type your answers in interval notation. use a comma to separate answers as needed.)
○ c. the function f is decreasing on the open interval(s) and never increasing
(type your answer in interval notation. use a comma to separate answers as needed.)

Explanation:

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

We have \(f^{\prime}(x)=\frac{x^{2}(x - 2)}{x + 5}\), \(x
eq-5\).
First, find the critical points from \(f^{\prime}(x)=0\) (i.e., \(x^{2}(x - 2)=0\)) which gives \(x = 0\) or \(x=2\), and the point where \(f^{\prime}(x)\) is undefined \(x=-5\).
Now, consider the intervals \((-\infty,-5)\), \((-5,0)\), \((0,2)\) and \((2,\infty)\).
Take a test - point in each interval:

• For the interval \((-\infty,-5)\), let \(x=-6\). Then \(f^{\prime}(-6)=\frac{(-6)^{2}(-6 - 2)}{-6 + 5}=\frac{36\times(-8)}{-1}=288>0\).
• For the interval \((-5,0)\), let \(x=-1\). Then \(f^{\prime}(-1)=\frac{(-1)^{2}(-1 - 2)}{-1 + 5}=\frac{1\times(-3)}{4}=-\frac{3}{4}<0\).
• For the interval \((0,2)\), let \(x = 1\). Then \(f^{\prime}(1)=\frac{1^{2}(1 - 2)}{1+5}=\frac{1\times(-1)}{6}=-\frac{1}{6}<0\).
• For the interval \((2,\infty)\), let \(x = 3\). Then \(f^{\prime}(3)=\frac{3^{2}(3 - 2)}{3 + 5}=\frac{9\times1}{8}=\frac{9}{8}>0\).

Step2: Determine increasing and decreasing intervals

Since \(f^{\prime}(x)>0\) on \((-\infty,-5)\cup(2,\infty)\) and \(f^{\prime}(x)<0\) on \((-5,2)\) (but \(x
eq - 5\)).
The function \(y = f(x)\) is increasing when \(f^{\prime}(x)>0\) and decreasing when \(f^{\prime}(x)<0\).

Answer:

The function \(f\) is increasing on the open intervals \((-\infty,-5)\) and \((2,\infty)\) and decreasing on the open interval \((-5,2)\).