What is the critical f(x)=x^2ln(3x)+6 ?

The critical f(x)=x^2ln(3x)+6 is x= 1/(3sqrt(e))

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critical f(x)=x^2ln(3x)+6

critical points

f (x) = x^{2} ln (3 x) + 6

x = \frac{1}{3 e}

Solution steps

x^{2} ln (3 x) + 6

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{1}{3 e}, x = 0

Identify critical points not in

f (x)

domain

Domain of

x^{2} ln (3 x) + 6 : x > 0

x^{2} ln (3 x) + 6

is not defined at

x = 0

, therefore:

x = \frac{1}{3 e}

f (x) = \frac{2 x}{x ^{2} - 36}

\frac{x ^{3}}{3} - 2 x^{2} - 5 x

f (x) = \frac{8 x ( x ^{2} - 3 )}{( x ^{2} + 1 ) ^{3}}

f (x) = 8 x^{3} - 3 x

f (x) = 36 - x^{2}

What is the critical f(x)=x^2ln(3x)+6 ?
The critical f(x)=x^2ln(3x)+6 is x= 1/(3sqrt(e))