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

The extreme f(x)=3^2\sqrt[3]{x^2}-2x is Minimum(0,0),Maximum(27,27)

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

extreme points

Minimum (0, 0), Maximum (27, 27)

Solution steps

Find the critical points:

x = 0, x = 27

Domain of

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 27,

Decreasing

: 27 < x < \infty

Plug the extreme point

x = 0

into

Minimum (0, 0)

Plug the extreme point

x = 27

into

Maximum (27, 27)

Minimum (0, 0), Maximum (27, 27)

f (x) = 6 x^{\frac{2}{3}} - x, 0 \leq x \leq 216

x 1 - x^{2}, - 1 \leq x \leq 1

7 + 4 x^{2} - x^{4}

f (x) = 1.3 t e^{- 2.9 t}

y = x \cdot 1 + x^{2}

What is the extreme f(x)=3^2\sqrt[3]{x^2}-2x ?
The extreme f(x)=3^2\sqrt[3]{x^2}-2x is Minimum(0,0),Maximum(27,27)