What is the extreme f(x)=-x^3+6x^2-9x+1 ?

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

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extreme f(x)=-x^3+6x^2-9x+1

extreme points

f (x) = - x^{3} + 6 x^{2} - 9 x + 1

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

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 + 1 : - 3

Minimum (1, - 3)

Plug

x = 3

into

- x^{3} + 6 x^{2} - 9 x + 1 : 1

Maximum (3, 1)

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

f (x, y) = 2 x - 4 y + 7

f (x, y) = x e^{y}

f (x) = x^{\frac{1}{3}} (x - 4)

f (x, y) = 4 x^{2} y + 2 x y^{2} - 12 x y - 5

e^{2 x} + e^{- x}

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