# calculus

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(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
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?

(c) take f''(x, set numerator equal to zero and solve but how?

• calculus -

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

• calculus-another Q -

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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