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

The critical f(x)=x(ln(x))^2 is x= 1/(e^2),x=1

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critical f(x)=x(ln(x))^2

critical points

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

x = \frac{1}{e ^{2}}, x = 1

Solution steps

x ln^{2} (x)

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{1}{e ^{2}}, x = 1, x = 0

Identify critical points not in

f (x)

domain

Domain of

x ln^{2} (x) : x > 0

x ln^{2} (x)

is not defined at

x = 0

, therefore:

x = \frac{1}{e ^{2}}, x = 1

x^{3} + y^{3} - 12 x - 3 y

f (x) = x^{\frac{6}{5}} - 12 x^{\frac{1}{5}}

f (x) = cos (π x)

What is the critical f(x)=x(ln(x))^2 ?
The critical f(x)=x(ln(x))^2 is x= 1/(e^2),x=1