What is the critical f(x)=2xln(16x^2) ?

The critical f(x)=2xln(16x^2) is x=-1/(4e),x= 1/(4e)

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critical f(x)=2xln(16x^2)

critical points

f (x) = 2 x ln (16 x^{2})

x = - \frac{1}{4 e}, x = \frac{1}{4 e}

Solution steps

2 x ln (16 x^{2})

Find where

f^{'} (x)

is equal to zero or undefined

x = - \frac{1}{4 e}, x = \frac{1}{4 e}, x = 0

Identify critical points not in

f (x)

domain

Domain of

2 x ln (16 x^{2}) : x < 0 or x > 0

2 x ln (16 x^{2})

is not defined at

x = 0

, therefore:

x = - \frac{1}{4 e}, x = \frac{1}{4 e}

\frac{3 x ^{2}}{x ^{2} - 4}

f (x) = x^{4} - 3 x^{2} + 2

f (x) = \frac{x}{x ^{2} + 81}

f (x, y) = (x - 1)^{2} + (x - y)^{4}

x^{2} - 6 x + 7

What is the critical f(x)=2xln(16x^2) ?
The critical f(x)=2xln(16x^2) is x=-1/(4e),x= 1/(4e)