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

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

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

extreme points

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

Minimum (0, 0), Maximum (4, 32)

Solution steps

f^{'} (x) = - 3 x^{2} + 12 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 4,

Decreasing

: 4 < x < \infty

Plug

x = 0

into

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

Minimum (0, 0)

Plug

x = 4

into

- x^{3} + 6 x^{2} : 32

Maximum (4, 32)

Minimum (0, 0), Maximum (4, 32)

f (x) = 3 x - 24

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

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

f (x) = 5 + 4 x - x^{3}

y = (x + 2)^{2}

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