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) ).\nfind horizontal asymptote(s), if any. select the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. the horizontal asymptote(s) is (are) ( y= ) (use a comma to separate answers as needed.)\nb. there are no horizontal asymptotes.\nsummarize the pertinent information obtained by analyzing ( f^{prime}(x) ). select the correct choice below and fill in the answer box(es) to complete your choice. (type your ans use a comma to separate answers as needed.)\na. ( f(x) ) is in d decreasing on ( (-infty, 0) ).\nb. ( f(x) ) is in minimum.\nc. ( f(x) ) is di maximum.\n( f(x) ) has a local

summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of ( f(x)=ln left(x^{2}+36\right) ).\nfind horizontal asymptote(s), if any. select the correct choice below and, if necessary, fill in the answer box to complete your choice.\na. the horizontal asymptote(s) is (are) ( y= ) (use a comma to separate answers as needed.)\nb. there are no horizontal asymptotes.\nsummarize the pertinent information obtained by analyzing ( f^{prime}(x) ). select the correct choice below and fill in the answer box(es) to complete your choice. (type your ans use a comma to separate answers as needed.)\na. ( f(x) ) is in d decreasing on ( (-infty, 0) ).\nb. ( f(x) ) is in minimum.\nc. ( f(x) ) is di maximum.\n( f(x) ) has a local

Answer

Explanation:

Step1: Find the derivative of ( f(x)=\ln(x^{2}+36) )

Using the chain rule, if ( y = \ln(u) ) and ( u=x^{2}+36 ), then ( y^\prime=\frac{u^\prime}{u} ). Since ( u^\prime = 2x ), we have ( f^\prime(x)=\frac{2x}{x^{2}+36} ).

Step2: Analyze the sign of ( f^\prime(x) )

Set ( f^\prime(x)=0 ), then ( \frac{2x}{x^{2}+36}=0 ). Since ( x^{2}+36>0 ) for all real ( x ), the critical point is ( x = 0 ). For ( x>0 ), ( f^\prime(x)=\frac{2x}{x^{2}+36}>0 ) (because ( 2x>0 ) and ( x^{2}+36>0 )). For ( x<0 ), ( f^\prime(x)=\frac{2x}{x^{2}+36}<0 ) (because ( 2x<0 ) and ( x^{2}+36>0 )).

Answer:

A. ( f(x) ) is increasing on ( (0,\infty) ) and decreasing on ( (-\infty,0) ). There is a local minimum at ( x = 0 ). For the horizontal - asymptote part: We know that ( \lim_{x\rightarrow\pm\infty}\ln(x^{2}+36)=\infty ). So, the answer for the horizontal - asymptote is B. There are no horizontal asymptotes.