Graph ( natural log of x)/(x^2)

Problem

ƒ(x)=ln(x)/(x2)

Solution

1. Identify the domain of the function. Since the natural logarithm ln(x) is only defined for positive values, the domain is x>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 vertical asymptotes by checking the limit as x approaches the boundary of the domain.

(lim_x→0)(ln(x)/(x2))=−∞

There is a vertical asymptote at x=0

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

(lim_x→∞)(ln(x)/(x2))=(lim_x→∞)(1/x/(2*x))

(lim_x→∞)(1/(2*x2))=0

There is a horizontal asymptote at y=0

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

ƒ(x)′=(x2⋅1/x−ln(x)⋅2*x)/((x2)2)

ƒ(x)′=(x−2*x*ln(x))/(x4)

ƒ(x)′=(1−2*ln(x))/(x3)

6. Solve for relative extrema by setting ƒ(x)′=0

1−2*ln(x)=0

ln(x)=1/2

x=e(1/2)=√(,e)≈1.649

The maximum value is ƒ(√(,e))=(1/2)/e=1/(2*e)≈0.184

7. Find inflection points by calculating the second derivative.

ƒ(x)″=(x3⋅(−2)/x−(1−2*ln(x))⋅3*x2)/(x6)

ƒ(x)″=(−2*x2−3*x2+6*x2*ln(x))/(x6)

ƒ(x)″=(6*ln(x)−5)/(x4)

Setting ƒ(x)″=0 gives ln(x)=5/6 so x=e(5/6)≈2.301

Final Answer

The graph of ƒ(x)=ln(x)/(x2) has a domain of (0,∞) an x-intercept at (1,0) a vertical asymptote at x=0 a horizontal asymptote at y=0 a local maximum at (√(,e),1/(2*e)) and an inflection point at (e(5/6),5/(6*e(5/3)))

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