Question

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

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 \( \lim_{x
ightarrow\pm\infty}\ln(x^{2}+36)=\infty \). So, the answer for the horizontal - asymptote is B. There are no horizontal asymptotes.