What is the extreme f(x)=-3x^2+x^3 ?

The extreme f(x)=-3x^2+x^3 is Maximum(0,0),Minimum(2,-4)

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extreme f(x)=-3x^2+x^3

extreme points

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

Maximum (0, 0), Minimum (2, - 4)

Solution steps

f^{'} (x) = - 6 x + 3 x^{2}

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

- 3 x^{2} + x^{3} : 0

Maximum (0, 0)

Plug

x = 2

into

- 3 x^{2} + x^{3} : - 4

Minimum (2, - 4)

Maximum (0, 0), Minimum (2, - 4)

f (x) = 4 x e^{3 x}

f (x) = 6 x^{\frac{1}{6}} y^{\frac{1}{6}} - x - y

(x - 4) e^{x}

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

f (x, y) = x^{3} + 2 x y^{2} - \frac{y ^{4}}{x}

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