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

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

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

extreme points

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

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

Solution steps

Find the critical points:

x = 0, x = 1

Domain of

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug the extreme point

x = 0

into

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

Maximum (0, 0)

Plug the extreme point

x = 1

into

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

Minimum (1, - 1)

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

f (x, y) = 4 x + 6 y - x^{2} - y^{2}

f (x) = x^{2} + 4 x + 16

f (x) = - 2 x^{2} + 4 x - 4

f (x) = - 2 x^{2} + 4 x + 4

\frac{x ^{3} - 3 x}{x ^{2} - 1}

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