What is the critical f(x)=(1-ln(x))/(x^2) ?

The critical f(x)=(1-ln(x))/(x^2) is x=e^{3/2}

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critical f(x)=(1-ln(x))/(x^2)

critical points

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

x = e^{\frac{3}{2}}

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = e^{\frac{3}{2}}, x = 0

Identify critical points not in

f (x)

domain

Domain of

\frac{1 - ln ( x )}{x ^{2}} : x > 0

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

is not defined at

x = 0

, therefore:

x = e^{\frac{3}{2}}

f (x) = sin (x) + cos (x), (0, 2 π)

f (x) = sin (\frac{x}{2})

f (x, y) = 3 x^{3} y + y^{2}

f (x, y) = 3 x^{2} - 2 x y + y^{2} - 8 y

f (x, y) = x^{2} - 2 x y + 2 y^{2} - 2 y

What is the critical f(x)=(1-ln(x))/(x^2) ?
The critical f(x)=(1-ln(x))/(x^2) is x=e^{3/2}