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

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### 2 Answers

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))**

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)).**