What is the critical f(x)=ln(8+27x^3) ?

The critical f(x)=ln(8+27x^3) is x=0

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critical f(x)=ln(8+27x^3)

critical points

f (x) = ln (8 + 27 x^{3})

x = 0

Solution steps

ln (8 + 27 x^{3})

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = - \frac{2}{3}

Identify critical points not in

f (x)

domain

Domain of

ln (8 + 27 x^{3}) : x > - \frac{2}{3}

ln (8 + 27 x^{3})

is not defined at

x = - \frac{2}{3}

, therefore:

x = 0

f (x) = sin^{2} (5 x) + cos (5 x), 0 < x < \frac{2 π}{5}

f (x) = 3 x + 5 x^{- 1}

(x^{2} - 1)^{\frac{2}{3}}

4 x^{2}

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