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# FunctionsDetermine the local maxim value of f(x) = ln x / (6*sqrt x).

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Determine the local maxim value of f(x) = ln x / (6*sqrt x).

Posted by l0l1 on May 19, 2011 at 6:48 AM via web and tagged with algebra1, discussion

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To find the local extreme of a function, first we have to determine the critical value of the function.

The critical value of the function is the root of the first derivative of the function. We'll differentiate the function and we'll get:

f'(x) = [ln x / (6*sqrt x)]'

We'll apply the quotient rule:

f'(x) = [(ln x )'* (6*sqrt x) - (ln x)*(6*sqrt x)]/36x

f'(x) = [(6*sqrt x/x) - 6lnx/2sqrtx]/36x

f'(x) = (18x - 6x*lnx)/72x^2*sqrt x

We'll factorize by 6x:

f'(x) = 6x(3 - lnx)/72x^2*sqrt x

We'll simplify by 6x:

f'(x) = (3 - lnx)/12x*sqrt x

f'(x) = 0

3 - lnx = 0

ln x = 3

x = e^3

The critical value for the given function f(x) is x = e^3.

The local extreme is f( e^3);

f( e^3) = [ln e^3/ (6*sqrt e^3)]

f( e^3) = 3ln e/6e*sqrt e, where ln e = 1

f( e^3) = 1/2e*sqrt e

f( e^3) = 0.10787 (the considered value for e = 2.78 approx.)

Posted by giorgiana1976 on May 19, 2011 at 11:20 AM (Answer #2)

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