What is the extreme x\sqrt[3]{x^2-9} ?

The extreme x\sqrt[3]{x^2-9} is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme x\sqrt[3]{x^2-9}

extreme

Maximum (- \frac{3 3}{5}, \frac{3 3 18 3}{5 ^{\frac{5}{6}}}), Minimum (\frac{3 3}{5}, - \frac{3 3 18 3}{5 ^{\frac{5}{6}}})

Solution steps

Find the critical points:

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

Domain of

x 3 x^{2} - 9 : - \infty < x < \infty

Find intervals:

Increasing

: - \infty < x < - 3,

Increasing

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

Decreasing

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

Increasing

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

Increasing

: 3 < x < \infty

Plug the extreme point

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

into

x 3 x^{2} - 9 \Rightarrow y = \frac{3 3 18 3}{5 ^{\frac{5}{6}}}

Maximum (- \frac{3 3}{5}, \frac{3 3 18 3}{5 ^{\frac{5}{6}}})

Plug the extreme point

x = \frac{3 3}{5}

into

x 3 x^{2} - 9 \Rightarrow y = - \frac{3 3 18 3}{5 ^{\frac{5}{6}}}

Minimum (\frac{3 3}{5}, - \frac{3 3 18 3}{5 ^{\frac{5}{6}}})

Maximum (- \frac{3 3}{5}, \frac{3 3 18 3}{5 ^{\frac{5}{6}}}), Minimum (\frac{3 3}{5}, - \frac{3 3 18 3}{5 ^{\frac{5}{6}}})