Graph ( natural log of x)/x

Problem

ƒ(x)=ln(x)/x

Solution

1. Identify the domain of the function. Since the natural logarithm ln(x) is only defined for positive values and the denominator x cannot be zero, the domain is (0,∞)

2. Find the x-intercept by setting the numerator to zero.

ln(x)=0

x=e0=1

The graph crosses the x-axis at (1,0)

3. Determine the vertical asymptote by checking the limit as x approaches 0 from the right.

(lim_x→0)(ln(x)/x)=−∞

There is a vertical asymptote at x=0

4. Determine the horizontal asymptote by checking the limit as x approaches infinity using L'Hôpital's Rule.

(lim_x→∞)(ln(x)/x)=(lim_x→∞)(1/x/1)=0

There is a horizontal asymptote at y=0

5. Find the critical points by calculating the first derivative using the quotient rule.

ƒ(x)′=(x⋅1/x−ln(x)⋅1)/(x2)

ƒ(x)′=(1−ln(x))/(x2)

Setting ƒ(x)′=0 gives 1−ln(x)=0 so x=e

6. Determine the local maximum by evaluating the function at the critical point.

ƒ(e)=ln(e)/e=1/e≈0.368

The function increases on (0,e) and decreases on (e,∞) making (e,1/e) a local maximum.

7. Find the inflection point by calculating the second derivative.

ƒ(x)″=(x2*(−1/x)−(1−ln(x))*(2*x))/(x4)

ƒ(x)″=(−x−2*x+2*x*ln(x))/(x4)

ƒ(x)″=(2*ln(x)−3)/(x3)

Setting ƒ(x)″=0 gives 2*ln(x)=3 so x=e1.5≈4.48

Final Answer

The graph of ƒ(x)=ln(x)/x starts from a vertical asymptote at x=0 passes through (1,0) reaches a local maximum at (e,1/e) and approaches the x-axis as x→∞

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