What are inflection points of f(x)=ln x/x, if x>0?
To find the points of inflection we need to find the second derivative of f(x) and find the points where the second derivative changes sign.
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The inflection points are the roots of the second derivative of the function.
We'll have to determine the 1st derivative, for the beginning. We'll do this using quotient rule:
f'(x) = [(ln x)'*x - (ln x)*x']/x^2
f'(x) = (x/x - ln x)/x^2
f'(x) = (1 - ln x)/x^2
Now, we'll determine the 2nd derivative using quotient rule:
f"(x) = [(1 - ln x)'*x^2 - (1 - ln x)*(x^2)']/x^4
f"(x) = [(-1/x)*x^2 - 2x*(1 - ln x)]/x^4
f"(x) = [-x - 2x + (2x*ln x)]/x^4
f"(x) = (2ln x - 3)/x^3
We'll equate and we'll get:
f"(x) = 0
(2ln x - 3)/x^3 = 0 <=> 2ln x - 3 = 0 => 2ln x = 3 => ln x = 3/2
x = e^(3/2)
x = sqrt e^3
f(e^(3/2)) = [ln e^(3/2)]/e^(3/2)
We'll use the power property of logarithms:
f(e^(3/2)) = 3/2e^(3/2)
f(e^(3/2)) = 3/(2e*sqrt e)
The inflection point has the coordinates: (e^(3/2) ; 3/(2e*sqrt e)).