What is the extreme f(x)=x^3-9x^2+8 ?

The extreme f(x)=x^3-9x^2+8 is Maximum(0,8),Minimum(6,-100)

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extreme f(x)=x^3-9x^2+8

extreme points

f (x) = x^{3} - 9 x^{2} + 8

Maximum (0, 8), Minimum (6, - 100)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = 0

into

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

Maximum (0, 8)

Plug

x = 6

into

x^{3} - 9 x^{2} + 8 : - 100

Minimum (6, - 100)

Maximum (0, 8), Minimum (6, - 100)

f (x) = \frac{- 6}{x - 7}

f (x) = x ln (\frac{x}{5})

f (x) = x^{3} - 2 x^{2} - 4 x + 12

f (x) = - 5 cos (3 x) - 2

f (x) = x^{3} - 9 x^{2} + 24 x - 12

What is the extreme f(x)=x^3-9x^2+8 ?
The extreme f(x)=x^3-9x^2+8 is Maximum(0,8),Minimum(6,-100)