Posted by Riley on Monday, April 6, 2009 at 9:44am.
(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.
f(x)=(lnx)/sqrtx
I took the derivative and got
2lnx/2x^3/2
take the values when f'(x) >0 I get
2lnx > 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?

calculus  Reiny, Monday, April 6, 2009 at 12:23pm
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

calculusanother Q  Riley, Monday, April 6, 2009 at 1:55pm
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.
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