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

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)

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)