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

The extreme f(x)=3x-2x^2-(4x^3)/3 is Minimum(-3/2 ,-9/2),Maximum(1/2 , 5/6)

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

extreme points

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

Minimum (- \frac{3}{2}, - \frac{9}{2}), Maximum (\frac{1}{2}, \frac{5}{6})

Solution steps

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

Find intervals:

Decreasing

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

Increasing

: - \frac{3}{2} < x < \frac{1}{2},

Decreasing

: \frac{1}{2} < x < \infty

Plug

x = - \frac{3}{2}

into

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

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

Plug

x = \frac{1}{2}

into

3 x - 2 x^{2} - \frac{4 x ^{3}}{3} : \frac{5}{6}

Maximum (\frac{1}{2}, \frac{5}{6})

Minimum (- \frac{3}{2}, - \frac{9}{2}), Maximum (\frac{1}{2}, \frac{5}{6})

f (x, y) = 2 x^{2} + 3 y^{2} - 7

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

f (x, y) = x^{3} - 6 x y + 3 y^{2} + 1

f (x) = \frac{x ^{2}}{x - 6}

f (x, y) = 4 x^{3} - 11 x y^{2} + 15 y + 12

What is the extreme f(x)=3x-2x^2-(4x^3)/3 ?
The extreme f(x)=3x-2x^2-(4x^3)/3 is Minimum(-3/2 ,-9/2),Maximum(1/2 , 5/6)