What is the extreme f(x)=-3x^3+6x^2 ?

The extreme f(x)=-3x^3+6x^2 is Minimum(0,0),Maximum(4/3 , 32/9)

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extreme f(x)=-3x^3+6x^2

extreme points

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

Minimum (0, 0), Maximum (\frac{4}{3}, \frac{32}{9})

Solution steps

f^{'} (x) = - 9 x^{2} + 12 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{4}{3},

Decreasing

: \frac{4}{3} < x < \infty

Plug

x = 0

into

- 3 x^{3} + 6 x^{2} : 0

Minimum (0, 0)

Plug

x = \frac{4}{3}

into

- 3 x^{3} + 6 x^{2} : \frac{32}{9}

Maximum (\frac{4}{3}, \frac{32}{9})

Minimum (0, 0), Maximum (\frac{4}{3}, \frac{32}{9})

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

f (x) = 4 x^{3} - 9 x^{2} + 3 x + 1

x^{3} + 2 x y + y^{2} - 5 x

\frac{- 4}{x ^{2} - 9}

f (x) = e^{x^{2} - 2 x - 1}

What is the extreme f(x)=-3x^3+6x^2 ?
The extreme f(x)=-3x^3+6x^2 is Minimum(0,0),Maximum(4/3 , 32/9)