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October 30, 2014

October 30, 2014

Posted by **Riley** on Monday, April 6, 2009 at 9:44am.

(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

2-lnx/2x^3/2

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?

- calculus -
**Reiny**, Monday, April 6, 2009 at 12:23pmyou 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

- calculus-another Q -
**Riley**, Monday, April 6, 2009 at 1:55pmokay 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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