What is the extreme f(x)=x^{-2}ln(x) ?

The extreme f(x)=x^{-2}ln(x) is Maximum(sqrt(e), 1/(2e))

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extreme f(x)=x^{-2}ln(x)

extreme points

f (x) = x^{- 2} ln (x)

Maximum (e, \frac{1}{2 e})

Solution steps

f^{'} (x) = \frac{- 2 ln ( x ) + 1}{x ^{3}}

Find intervals:

Increasing

: 0 < x < e,

Decreasing

: e < x < \infty

Plug

x = e

into

x^{- 2} ln (x) : \frac{1}{2 e}

Maximum (e, \frac{1}{2 e})

x^{4} - 8 x^{2} + 9

f (x) = \frac{1}{1 - x ^{2} - y ^{2}}

5 x^{2} ln (x)

f (x) = (6 x - x^{2})

f (x, y) = \frac{2 x y}{( x - 2 )}

What is the extreme f(x)=x^{-2}ln(x) ?
The extreme f(x)=x^{-2}ln(x) is Maximum(sqrt(e), 1/(2e))