What is the extreme h(x)=-6x^3+18x^2+3 ?

The extreme h(x)=-6x^3+18x^2+3 is Minimum(0,3),Maximum(2,27)

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extreme h(x)=-6x^3+18x^2+3

extreme points

h (x) = - 6 x^{3} + 18 x^{2} + 3

Minimum (0, 3), Maximum (2, 27)

Solution steps

f^{'} (x) = - 18 x^{2} + 36 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = 0

into

- 6 x^{3} + 18 x^{2} + 3 : 3

Minimum (0, 3)

Plug

x = 2

into

- 6 x^{3} + 18 x^{2} + 3 : 27

Maximum (2, 27)

Minimum (0, 3), Maximum (2, 27)

F (x) = \frac{2 x}{x ^{2} + 16}, - 8 \leq x \leq 8

- (\frac{9}{2})^{2} - 7^{2} + \frac{9}{2} - 14 + 3

x (8 - x^{2})

f (x) = f (x) = - 8 x^{2} - 144 x + 6

f (x) = x^{3} - 3 x [0.3]

What is the extreme h(x)=-6x^3+18x^2+3 ?
The extreme h(x)=-6x^3+18x^2+3 is Minimum(0,3),Maximum(2,27)