Question

summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of ( f(x)=\frac{x + 2}{x - 2} )
(type an integer or a decimal. use a comma to separate answers as needed.)
there is a local maximum at ( x= ). there is no local minimum.
(type an integer or a decimal. use a comma to separate answers as needed.)
there is a local maximum at ( x= ) and there is a local minimum at ( x= ).
(type integers or decimals. use a comma to separate answers as needed.)
there are no local extrema.
find the intervals where ( f(x) ) is concave upward or downward. select the correct choice below and fill in the answer box(es) to complete your choice.
(type your answer in interval notation. use a comma to separate answers as needed)
the function is concave upward on. it is never concave downward.
the function is concave downward on. it is never concave upward.
the function is concave upward on. it is concave downward on.

Explanation:

Step1: Find the first derivative

Use the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\). Here \(u = x + 2\), \(u^\prime=1\), \(v=x - 2\), \(v^\prime = 1\).

$$
LATEXBLOCK0
$$

Since \(f^\prime(x)=\frac{-4}{(x - 2)^{2}}<0\) for all \(x
eq2\), there are no critical points (where \(f^\prime(x) = 0\) or \(f^\prime(x)\) is undefined, but at \(x = 2\) the function is not defined). So, there are no local extrema.

Step2: Find the second derivative

Use the quotient rule again. Let \(u=-4\), \(u^\prime = 0\), \(v=(x - 2)^{2}\), \(v^\prime=2(x - 2)\)

$$
LATEXBLOCK1
$$

Set \(f^{\prime\prime}(x)=0\), there is no solution. Consider the sign of \(f^{\prime\prime}(x)\):
When \(x>2\), \(f^{\prime\prime}(x)>0\), the function is concave upward on \((2,\infty)\)
When \(x<2\), \(f^{\prime\prime}(x)<0\), the function is concave downward on \((-\infty,2)\)

Answer:

• For local extrema: D. There are no local extrema.
• For concavity: C. The function is concave upward on \((2,\infty)\). It is concave downward on \((-\infty,2)\)