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

The extreme x\sqrt[3]{x^2-4} is

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

extreme points

x 3 x^{2} - 4

Maximum (- \frac{2 3}{5}, \frac{4 3}{5 ^{\frac{5}{6}}}), Minimum (\frac{2 3}{5}, - \frac{4 3}{5 ^{\frac{5}{6}}})

Solution steps

Find the critical points:

x = - 2, x = - \frac{2 3}{5}, x = \frac{2 3}{5}, x = 2

Domain of

x 3 x^{2} - 4 : - \infty < x < \infty

Find intervals:

Increasing

: - \infty < x < - 2,

Increasing

: - 2 < x < - \frac{2 3}{5},

Decreasing

: - \frac{2 3}{5} < x < \frac{2 3}{5},

Increasing

: \frac{2 3}{5} < x < 2,

Increasing

: 2 < x < \infty

Plug the extreme point

x = - \frac{2 3}{5}

into

x 3 x^{2} - 4 \Rightarrow y = \frac{4 3}{5 ^{\frac{5}{6}}}

Maximum (- \frac{2 3}{5}, \frac{4 3}{5 ^{\frac{5}{6}}})

Plug the extreme point

x = \frac{2 3}{5}

into

x 3 x^{2} - 4 \Rightarrow y = - \frac{4 3}{5 ^{\frac{5}{6}}}

Minimum (\frac{2 3}{5}, - \frac{4 3}{5 ^{\frac{5}{6}}})

Maximum (- \frac{2 3}{5}, \frac{4 3}{5 ^{\frac{5}{6}}}), Minimum (\frac{2 3}{5}, - \frac{4 3}{5 ^{\frac{5}{6}}})

3 x^{3} + 8

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

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

f (x) = x ln (\frac{x}{2})

f (x) = \frac{1}{2} (3 x - 1)