What is the extreme f(x)=3\sqrt[3]{x^2}-2x ?

The extreme f(x)=3\sqrt[3]{x^2}-2x is Minimum(0,0),Maximum(1,1)

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extreme f(x)=3\sqrt[3]{x^2}-2x

extreme points

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

Minimum (0, 0), Maximum (1, 1)

Solution steps

Find the critical points:

x = 0, x = 1

Domain of

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < \infty

Plug the extreme point

x = 0

into

3 3 x^{2} - 2 x \Rightarrow y = 0

Minimum (0, 0)

Plug the extreme point

x = 1

into

3 3 x^{2} - 2 x \Rightarrow y = 1

Maximum (1, 1)

Minimum (0, 0), Maximum (1, 1)

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

f (x) = 2 x + \frac{20}{x}

12 x - 4 x^{2}

f (x, y) = x^{3} + y^{2} - 6 x y - 29 x + 18 y + 20

f (x) = 4 + 81 x - 3 x^{3}, 0 \leq x \leq 4

What is the extreme f(x)=3\sqrt[3]{x^2}-2x ?
The extreme f(x)=3\sqrt[3]{x^2}-2x is Minimum(0,0),Maximum(1,1)