What are inflection points of f(x)=ln x/x, if x>0?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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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.

f(x) = ln x / x

f'(x) = (1/x)*(1/x) - (ln x) / x^2

f''(x) = -2*x^-3 - [ (1/x^2)(1/x) - 2*ln x / x^3]

=> -2x^-3 - 1/ x^3 + 2*ln x / x^3

=> (-3 + 2*ln x) / x^3

Equating f''(x) = 0

=> 2*ln x = 3

=> ln x = 3/2

x = e^(3/2)

f(e^(3/2)) = ln e^3/2 / e^(3/2)

=> (3/2)/e^(3/2)

The point of inflection is ( e^(3/2) ,  (3/2)/e^(3/2))

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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)).

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