summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of (…

summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of ( f(x)=ln left(x^{2}+36\right) ).\n(type your answer in interval notation. use a comma to separate answers as needed.)\na. ( f(x) ) is increasing on ( (0, infty) ) and decreasing on ( (-infty, 0) ).\nb. ( f(x) ) is increasing on\nc. ( f(x) ) is decreasing on\n( f(x) ) has a local minimum.\nsummarize the pertinent information obtained by analyzing ( f^{prime prime}(x) ). select the correct choice below and fill in the answer box(es) to complete your choice.\n(type your answer in interval notation. use a comma to separate answers as needed.)\na. ( f(x) ) is concave upward on and concave downward on\nb. ( f(x) ) is concave upward on\nc. ( f(x) ) is concave downward on

summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of ( f(x)=ln left(x^{2}+36\right) ).\n(type your answer in interval notation. use a comma to separate answers as needed.)\na. ( f(x) ) is increasing on ( (0, infty) ) and decreasing on ( (-infty, 0) ).\nb. ( f(x) ) is increasing on\nc. ( f(x) ) is decreasing on\n( f(x) ) has a local minimum.\nsummarize the pertinent information obtained by analyzing ( f^{prime prime}(x) ). select the correct choice below and fill in the answer box(es) to complete your choice.\n(type your answer in interval notation. use a comma to separate answers as needed.)\na. ( f(x) ) is concave upward on and concave downward on\nb. ( f(x) ) is concave upward on\nc. ( f(x) ) is concave downward on

Answer

Explanation:

Step1: Find the first - derivative

We know that if (y = \ln(u)), then (y^\prime=\frac{u^\prime}{u}). Let (u=x^{2}+36), then (u^\prime = 2x). So (f^\prime(x)=\frac{2x}{x^{2}+36}). Set (f^\prime(x)=0), we get (2x = 0), so (x = 0). When (x>0), (f^\prime(x)=\frac{2x}{x^{2}+36}>0); when (x < 0), (f^\prime(x)=\frac{2x}{x^{2}+36}<0).

Step2: Find the second - derivative

Using the quotient rule ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}), where (u = 2x), (u^\prime=2), (v=x^{2}+36), (v^\prime = 2x). (f^{\prime\prime}(x)=\frac{2(x^{2}+36)-2x\times(2x)}{(x^{2}+36)^{2}}=\frac{72 - 2x^{2}}{(x^{2}+36)^{2}}). Set (f^{\prime\prime}(x)=0), then (72-2x^{2}=0), (x^{2}=36), (x=\pm6). When (x\in(-6,6)), (f^{\prime\prime}(x)=\frac{72 - 2x^{2}}{(x^{2}+36)^{2}}>0); when (x\in(-\infty,-6)\cup(6,\infty)), (f^{\prime\prime}(x)=\frac{72 - 2x^{2}}{(x^{2}+36)^{2}}<0).

Answer:

A. (f(x)) is concave upward on ((-6,6)) and concave downward on ((-\infty,-6),(6,\infty))