Locate extrema and inflection points for `y=(ln(x))/x ` :
The domain is x>0.
Extrema can only occur at critical points, or points where the first derivative is zero or fails to exist.
`y=1/x*ln(x) `
`y'=1/x*(1/x)+(-1/x^2)lnx `
`y'=1/x^2(1-lnx) `
Setting equal to zero we get:
`1/x^2=(lnx)/x^2 ==> lnx=1 ==> x=e `
For 0<x<e the first derivative is positive and for x>e it is negative, so there is a maximum at x=e and this is the only extrema.
Inflection points can only occur if the second derivative is zero:
`y''=-2/(x^3)(1-lnx)+1/x^2(-1/x) `
`y''=-1/x^3(2-2lnx+1)=-1/x^3(3-2lnx) `
So `3-2lnx=0==> lnx=3/2 ==> x=e^(3/2)~~4.482 `
There is an inflection point at ` x=e^(1.5)` where the graph changes from concave down to concave up.
The graph:
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