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)

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

extreme points

\frac{4 x ^{2} - 3 x ^{4}}{2 x}

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})

f (x) = 2 x^{3} + 3 x^{2} - 36 x - 54

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

f (x) = - 3 sin (x) cos (x)

f (x) = \frac{4 x}{x ^{2} + 1}, (- 4, 0)

f (x) = x^{2} + y^{2} - 6 y + 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)