What is the extreme (4x^2-3x^4)/(2x) ?

The extreme (4x^2-3x^4)/(2x) is Minimum(-2/3 ,-8/9),Maximum(2/3 , 8/9)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme (4x^2-3x^4)/(2x)

extreme

Minimum (- \frac{2}{3}, - \frac{8}{9}), Maximum (\frac{2}{3}, \frac{8}{9})

Solution steps

f^{'} (x) = \frac{1}{2} (- 9 x^{2} + 4)

Find intervals:

Decreasing

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

Increasing

: - \frac{2}{3} < x < 0,

Increasing

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

Decreasing

: \frac{2}{3} < x < \infty

Plug

x = - \frac{2}{3}

into

\frac{4 x ^{2} - 3 x ^{4}}{2 x} : - \frac{8}{9}

Minimum (- \frac{2}{3}, - \frac{8}{9})

Plug

x = \frac{2}{3}

into

\frac{4 x ^{2} - 3 x ^{4}}{2 x} : \frac{8}{9}

Maximum (\frac{2}{3}, \frac{8}{9})

Minimum (- \frac{2}{3}, - \frac{8}{9}), Maximum (\frac{2}{3}, \frac{8}{9})

What is the extreme (4x^2-3x^4)/(2x) ?
The extreme (4x^2-3x^4)/(2x) is Minimum(-2/3 ,-8/9),Maximum(2/3 , 8/9)