Please help me solve this Find the open intervals where the function f(x) = ln (
ID: 2827979 • Letter: P
Question
Please help me solve this
Find the open intervals where the function f(x) = ln (x2 + 36) is concave upward or concave downward. Find any inflection points. Select the correct choice below and fill in any answer boxes within your choice. The function has a point of inflection at . (Type an ordered pair. Use a comma to separate answers as needed.) The function does not have an inflection point. Select the correct choice below and fill in any answer boxes within your choice. The function is concave upward on the interval(s) .(Type your answer in interval notation. Use a comma to separate answers as needed.) The function is never concave upward. Select the correct choice below and fill in any answer boxes within your choice. The function is concave downward on the interval(s) . (Type your answer in interval notation. Use a comma to separate answers as needed.) The function is never concave downward.Explanation / Answer
f'(x)=[2x/(x^2+36)]
1)A-points of nflection,f'(x)=0
x=0,y=ln(36) implies (0,3.58)
2)A-concave up,f'(x)>0
x>0, (0,infinity)
3)A-concave down,f'(x)<0
x<0,(-infinity,0)