What is the extreme \sqrt[3]{(x^2-4)^2} ?

The extreme \sqrt[3]{(x^2-4)^2} is Minimum(-2,0),Maximum(0,2\sqrt[3]{2}),Minimum(2,0)

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extreme \sqrt[3]{(x^2-4)^2}

extreme points

Solution steps

f^{'} (x) = \frac{4 x ( x ^{2} - 4 )}{3 ( ( x ^{2} - 4 ) ^{2} ) ^{\frac{2}{3}}}

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

Minimum (- 2, 0)

Plug

x = 0

into

Plug

x = 2

into

Minimum (2, 0)

f (x) = 2 x^{3} - 24 x

- x - 5

\frac{3}{x} + 9

f (x) = \frac{8 x + 15}{x ^{2} + 5 x}

f (x) = \frac{( x + 2 )}{( 2 x + 1 )}

What is the extreme \sqrt[3]{(x^2-4)^2} ?
The extreme \sqrt[3]{(x^2-4)^2} is Minimum(-2,0),Maximum(0,2\sqrt[3]{2}),Minimum(2,0)