posted by Riley on .
(a) find the intervals on which f is incrs or decrs.
(b) find the local min/max values of f.
(c) find the intervals of concavity and the inflection points.
I took the derivative and got
take the values when f'(x) >0 I get
2-lnx > 0
2 > lnx
now how do get x by itself on one side?
(b) how do I got about this?
(c) take f''(x, set numerator equal to zero and solve but how?
you should use brackets to obtain the proper order of operation,
f '(x) = (2- lnx)/(2x^(3/2))
why don't we do part b) first, then
2 - lnx = 0
lnx = 2 , by definition lnx = 2 <----> e^2 = x
so e^2 = x = appr. 7.39
so when x=e^2 we have a max/min value of the function
f(e^2) = ln(e^2)/√(e^2) = 2/2.718 = appr. .736
okay I get it now! But how do I find (C)find the intervals of concavity and the inflection points.
I think I would take the 2nd derivative and set the numerator equal to zero and solve but i'm not quite sure how solve it.