What is the extreme f(x)=(3x^2-x^3)^{1/3} ?

The extreme f(x)=(3x^2-x^3)^{1/3} is Minimum(0,0),Maximum(2,4^{1/3})

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extreme f(x)=(3x^2-x^3)^{1/3}

extreme points

f (x) = (3 x^{2} - x^{3})^{\frac{1}{3}}

Minimum (0, 0), Maximum (2, 4^{\frac{1}{3}})

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = 0

into

(3 x^{2} - x^{3})^{\frac{1}{3}} : 0

Minimum (0, 0)

Plug

x = 2

into

(3 x^{2} - x^{3})^{\frac{1}{3}} : 4^{\frac{1}{3}}

Maximum (2, 4^{\frac{1}{3}})

Minimum (0, 0), Maximum (2, 4^{\frac{1}{3}})

x^{2} - 2 x - 15

f (x) = x ln (\frac{x}{3})

x^{2} - 2 x - 18

F (x, y) = 200 x + 150 y

f (x) = ∣ x + 1 ∣

What is the extreme f(x)=(3x^2-x^3)^{1/3} ?
The extreme f(x)=(3x^2-x^3)^{1/3} is Minimum(0,0),Maximum(2,4^{1/3})