Graph 1/(x natural log of x)

Problem

ƒ(x)=1/(x*ln(x))

Solution

1. Identify the domain of the function. The natural logarithm ln(x) requires x>0 Additionally, the denominator cannot be zero, so x≠0 and ln(x)≠0 Since ln(1)=0 we must have x≠1 The domain is (0,1)∪(1,∞)

2. Determine vertical asymptotes by checking where the denominator is zero. As x→1^+ ƒ(x)→∞ As x→1^− ƒ(x)→−∞ Thus, x=1 is a vertical asymptote. As x→0^+ ln(x)→−∞ so x*ln(x)→0^− making ƒ(x)→−∞

3. Determine horizontal asymptotes by checking the limit as x→∞ As x increases, both x and ln(x) increase without bound, so x*ln(x)→∞ Therefore, (lim_x→∞)(ƒ(x))=0 The line y=0 is a horizontal asymptote.

4. Find the first derivative to determine intervals of increase and decrease. Using the power rule and product rule:

d(ƒ(x))/d(x)=−1/((x*ln(x))^2)⋅(ln(x)+1)

5. Locate critical points by setting the derivative to zero. ln(x)+1=0 when ln(x)=−1 which gives x=e^(−1)≈0.368

6. Analyze the sign of the derivative to find extrema. For 0<x<e^(−1) ƒ(x)^′>0 (increasing). For e^(−1)<x<1 ƒ(x)^′<0 (decreasing). For x>1 ƒ(x)^′<0 (decreasing). There is a local maximum at x=e^(−1) with y=1/(e^(−1)*ln(e^(−1)))=−e≈−2.718

7. Sketch the graph based on these features. The graph rises from −∞ at x=0 to a local maximum at (e^(−1),−e) then drops to −∞ as x approaches 1 from the left. To the right of x=1 the graph starts at ∞ and decreases toward the xaxis as x→∞

Final Answer

ƒ(x)=1/(x*ln(x))

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