What is the extreme y=x^3-6x^2+9x-8 ?

The extreme y=x^3-6x^2+9x-8 is Maximum(1,-4),Minimum(3,-8)

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extreme y=x^3-6x^2+9x-8

extreme points

y = x^{3} - 6 x^{2} + 9 x - 8

Maximum (1, - 4), Minimum (3, - 8)

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 - 8 : - 4

Maximum (1, - 4)

Plug

x = 3

into

x^{3} - 6 x^{2} + 9 x - 8 : - 8

Minimum (3, - 8)

Maximum (1, - 4), Minimum (3, - 8)

f (x) = \frac{( 1 )}{( 3 x ^{(2)} - 3 x + 1 )}

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

f (x) = \frac{x ^{2} - 24}{x - 5}

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

f (x) = ln (x) - x

What is the extreme y=x^3-6x^2+9x-8 ?
The extreme y=x^3-6x^2+9x-8 is Maximum(1,-4),Minimum(3,-8)