What is the extreme f(x)=-2x^3+6x^2+12 ?

The extreme f(x)=-2x^3+6x^2+12 is Minimum(0,12),Maximum(2,20)

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extreme f(x)=-2x^3+6x^2+12

extreme points

f (x) = - 2 x^{3} + 6 x^{2} + 12

Minimum (0, 12), Maximum (2, 20)

Solution steps

f^{'} (x) = - 6 x^{2} + 12 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 2,

Decreasing

: 2 < x < \infty

Plug

x = 0

into

- 2 x^{3} + 6 x^{2} + 12 : 12

Minimum (0, 12)

Plug

x = 2

into

- 2 x^{3} + 6 x^{2} + 12 : 20

Maximum (2, 20)

Minimum (0, 12), Maximum (2, 20)

(x - 2)^{3} \cdot (x + 2)

f (x) = \frac{x + 2}{x ^{2} - 7 x - 18}

5 x y - 7 x^{2} + 3 x - 6 y + 2

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

y = x^{2} - 6 x + 10

What is the extreme f(x)=-2x^3+6x^2+12 ?
The extreme f(x)=-2x^3+6x^2+12 is Minimum(0,12),Maximum(2,20)