The Intervals of concavity and convexity of f(x)=lnxx are
A
Convex :(e32,∞)
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B
Convex : (0,e32)
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C
Concave :(e32,∞)
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D
Concave : (0,e32)
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Solution
The correct option is D Concave : (0,e32) f(x)=lnxx f′(x)=1−lnxx2 f′′(x)=2lnx−3x3 f′′(x)>0⇒ convex and f′′(x)<0⇒ concave.
So, convex in x∈(e32,∞)
Concave in x∈(0,e32)