Question

summarize the pertinent information obtained by applying the graphing strategy and sketch the graph of ( f(x)=ln left(x^{2}+36
ight) ).

b. there are no vertical asymptotes.
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 answer in interval notation. use a comma to separate answers as needed.)
a. ( f(x) ) is increasing on and decreasing on
b. ( f(x) ) is increasing on
c. ( f(x) ) is decreasing on

Explanation:

Step1: Find the first - derivative

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

Step2: Determine the intervals of increase and decrease

Set \(f^\prime(x)=0\), so \(\frac{2x}{x^{2}+36}=0\). Since \(x^{2}+36>0\) for all real \(x\), the critical point is \(x = 0\).
Test intervals:

• For \(x<0\) (e.g., \(x=-1\)), \(f^\prime(-1)=\frac{2\times(-1)}{(-1)^{2}+36}=-\frac{2}{37}<0\).
• For \(x>0\) (e.g., \(x = 1\)), \(f^\prime(1)=\frac{2\times1}{1^{2}+36}=\frac{2}{37}>0\).

Answer:

A. \(f(x)\) is increasing on \((0,\infty)\) and decreasing on \((-\infty,0)\)