Question

summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of ( f(x)=\frac{x + 2}{x - 2} ).
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 ( (2,infty) ). it is concave downward on ( (-infty,2) ).
find the location of any inflection points of ( f(x) ). select the correct choice below and, if necessary, fill in the answer box to complete your choice.
there is an inflection point at ( x = ). (type an integer or a decimal. use a comma to separate answers as needed.)
there are no inflection points

Explanation:

Step1: Find the first - derivative

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

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Step2: Find the second - derivative

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

$$
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$$

Step3: Determine concavity

Set \(f^{\prime\prime}(x)=0\), \(\frac{8}{(x - 2)^{3}}=0\) has no solution.

• When \(x>2\), \(f^{\prime\prime}(x)>0\), so the function is concave upward on \((2,\infty)\).
• When \(x<2\), \(f^{\prime\prime}(x)<0\), so the function is concave downward on \((-\infty,2)\).

Step4: Determine inflection points

Since \(f(x)\) is not defined at \(x = 2\) (vertical asymptote \(x=2\)), and \(f^{\prime\prime}(x)\) has no sign change at a point where \(f(x)\) is defined.

Answer:

For the concavity:

• The function is concave upward on \((2,\infty)\) and concave downward on \((-\infty,2)\) (Option C).

For inflection points:

• There are no inflection points (Option B).