What is the critical f(x)=ln(x^3-3x^2+4) ?

The critical f(x)=ln(x^3-3x^2+4) is x=0

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critical f(x)=ln(x^3-3x^2+4)

critical points

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

x = 0

Solution steps

ln (x^{3} - 3 x^{2} + 4)

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = - 1, x = 2

Identify critical points not in

f (x)

domain

Domain of

ln (x^{3} - 3 x^{2} + 4) : - 1 < x < 2 or x > 2

ln (x^{3} - 3 x^{2} + 4)

is not defined at

x = - 1, x = 2

, therefore:

x = 0

- 294 x + 2 x^{3} + 6 x y^{2} - 3 y^{3}

f (x, y) = \frac{1}{x} - \frac{64}{y} + x y

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

x^{\frac{2}{3}} (x^{2} + x - \frac{3}{2})

f (x, y) = x^{3} + y^{2} - 3 x^{2} + 10 y + 6

What is the critical f(x)=ln(x^3-3x^2+4) ?
The critical f(x)=ln(x^3-3x^2+4) is x=0