F(x)= (((x^2 −4x)^2 ))^(1/3) {: ((local maximum)),((absolut maximum)) } =?


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Question Number 163080 by tounghoungko last updated on 03/Jan/22
$F(x)= (((x^2 −4x)^2 ))^(1/3) {: ((local maximum)),((absolut maximum)) } =?$
$F (x) = \sqrt[3]{{(x^{2} - 4 x)}^{2}}$ $\begin{matrix} l o c a l m a x i m u m \\ a b s o l u t m a x i m u m \end{matrix}} = ?$
	Answered by mahdipoor last updated on 03/Jan/22
	$F ′=(2/3)(x^2 −4x)^((−1)/3) (2x−4)=0 or ∄ ⇒x=2 , 4 , 0 ⇒ { ((F (±∞)=∞ )),((F (2)=(2)^(4/3) LM (Local Max))),((F (4)=0 LM)),((F (0)=0 LM)) :}$
	$F^{'} = \frac{2}{3} {(x^{2} - 4 x)}^{\frac{- 1}{3}} (2 x - 4) = 0 o r ∄$ $\Rightarrow x = 2, 4, 0$ $\Rightarrow {\begin{cases} F (\pm \infty) = \infty \\ F (2) = {(2)}^{\frac{4}{3}} L M (L o c a l M a x) \\ F (4) = 0 L M \\ F (0) = 0 L M \end{cases}$
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