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

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

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

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 1,

Decreasing

: 1 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = 1

into

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

Maximum (1, 3)

Plug

x = 3

into

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

Minimum (3, - 1)

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

f (x) = x^{3} - 6 x^{2} + 9 x - 4

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

f (x) = \frac{( 3 x - 57 )}{( x - 85 ) ^{7}}

f (x, y) = 3 - 25 - (x + 1)^{2} - (y - 1)^{2}

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

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