We are given the function f(x) = (ln x) / (6(sqrt x)).
Now to find the local maxima of the function, we have to differentiate f(x).
f'(x) = (1- ln x)/ (6* x^1.5)
Equating f'(x) to 0.
(1- ln x)/ (6* x^1.5) = 0
=> 1- ln x = 0
=> ln x = 1
=> x = e
Also, f''(x) = [(ln x -1)/ 4*x^2.5] - 1/6*x^2.5
For x = e, f''(x) = -1/6*e^2.5 which is negative. Therefore the value of f(x) at x=e is the maxima value.
At x = e,
f (e) = (ln e) / (6*sqrt e)
= 1/ 6*sqrt e
Therefore the local maximum value is 1/ 6*sqrt e.
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.)