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

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

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

extreme points

Maximum (- 1, 2), Minimum (0, 0), Maximum (1, 2)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 1,

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = - 1

into

Maximum (- 1, 2)

Plug

x = 0

into

Minimum (0, 0)

Plug

x = 1

into

Maximum (1, 2)

Maximum (- 1, 2), Minimum (0, 0), Maximum (1, 2)

f (x) = 3 x^{4} + 4 x^{3} - 120 x^{2} + 120

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

f (x) = x - x y + 2 y

f (x) = 14 x^{2} - 7 x^{4}

x^{2} + x + 4

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