What is the extreme f(x)=-x^3+6x^2-9x ?

The extreme f(x)=-x^3+6x^2-9x is Minimum(1,-4),Maximum(3,0)

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extreme f(x)=-x^3+6x^2-9x

extreme points

f (x) = - x^{3} + 6 x^{2} - 9 x

Minimum (1, - 4), Maximum (3, 0)

Solution steps

f^{'} (x) = - 3 x^{2} + 12 x - 9

Find intervals:

Decreasing

: - \infty < x < 1,

Increasing

: 1 < x < 3,

Decreasing

: 3 < x < \infty

Plug

x = 1

into

- x^{3} + 6 x^{2} - 9 x : - 4

Minimum (1, - 4)

Plug

x = 3

into

- x^{3} + 6 x^{2} - 9 x : 0

Maximum (3, 0)

Minimum (1, - 4), Maximum (3, 0)

f (x) = x^{2} + 9 x + 5

f (x) = x^{4} - 8 x^{2} + a

7 \cdot 8^{x + 8} + 6

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

f (x) = x^{2} + 81

What is the extreme f(x)=-x^3+6x^2-9x ?
The extreme f(x)=-x^3+6x^2-9x is Minimum(1,-4),Maximum(3,0)