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

The extreme f(x)=-x^3+6x^2-18 is Minimum(0,-18),Maximum(4,14)

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

extreme points

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

Minimum (0, - 18), Maximum (4, 14)

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} - 18 : - 18

Minimum (0, - 18)

Plug

x = 4

into

- x^{3} + 6 x^{2} - 18 : 14

Maximum (4, 14)

Minimum (0, - 18), Maximum (4, 14)

f (x) = 7 x^{2} + 4

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

f (x) = 7 x^{\frac{3}{5}}

9 - 7 x

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

What is the extreme f(x)=-x^3+6x^2-18 ?
The extreme f(x)=-x^3+6x^2-18 is Minimum(0,-18),Maximum(4,14)