Question

summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of ( f(x)=\frac{3 x^{2}+1}{x^{2}-9} ).
(type an integer or a decimal. use a comma to separate answers as needed.)
c 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.)
d. 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.
a. the function is concave upward on . it is never concave downward.
(type your answer in interval notation. use a comma to separate answers as needed.)
b. the function is concave upward on . it is concave downward on .
(type your answers in interval notation. use a comma to separate answers as needed.)
c. the function is concave downward on . it is never concave upward.
(type your answer in interval notation. use a comma to separate answers as needed.)

Explanation:

Step1: Find the first derivative

Use the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\). Let \(u = 3x^{2}+1\), \(u^\prime=6x\); \(v=x^{2}-9\), \(v^\prime = 2x\). Then \(f^\prime(x)=\frac{6x(x^{2}-9)-2x(3x^{2}+1)}{(x^{2}-9)^{2}}=\frac{6x^{3}-54x - 6x^{3}-2x}{(x^{2}-9)^{2}}=\frac{- 56x}{(x^{2}-9)^{2}}\).
Set \(f^\prime(x)=0\), then \(-56x = 0\), so \(x = 0\).

Step2: Analyze the sign of the first derivative

When \(x\lt0\), \(f^\prime(x)\gt0\); when \(x\gt0\), \(f^\prime(x)\lt0\). So \(x = 0\) is a local maximum.

Step3: Find the second derivative

Use the quotient rule again. Let \(u=-56x\), \(u^\prime=-56\); \(v=(x^{2}-9)^{2}\), \(v^\prime = 2(x^{2}-9)\times2x=4x(x^{2}-9)\). Then \(f^{\prime\prime}(x)=\frac{-56(x^{2}-9)^{2}+56x\times4x(x^{2}-9)}{(x^{2}-9)^{4}}=\frac{-56(x^{2}-9)+224x^{2}}{(x^{2}-9)^{3}}=\frac{-56x^{2}+504 + 224x^{2}}{(x^{2}-9)^{3}}=\frac{168x^{2}+504}{(x^{2}-9)^{3}}=\frac{168(x^{2}+3)}{(x^{2}-9)^{3}}\).

Step4: Analyze the concavity

Set \(f^{\prime\prime}(x)=0\), since \(x^{2}+3\gt0\) for all real \(x\), the numerator \(168(x^{2}+3)\gt0\). The denominator \((x^{2}-9)^{3}\gt0\) when \(x^{2}-9\gt0\) (i.e., \(x\lt - 3\) or \(x\gt3\)) and \((x^{2}-9)^{3}\lt0\) when \(x^{2}-9\lt0\) (i.e., \(-3\lt x\lt3\)).
When \(x\lt - 3\) or \(x\gt3\), \(f^{\prime\prime}(x)\gt0\) (concave upward); when \(-3\lt x\lt3\), \(f^{\prime\prime}(x)\lt0\) (concave downward).

Answer:

C. There is a local maximum at \(x = 0\) and there is no local minimum.
B. The function is concave upward on \((-\infty,-3),(3,\infty)\). It is concave downward on \((-3,3)\).