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

The extreme f(x)=x^3-9x^2+9 is Maximum(0,9),Minimum(6,-99)

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

extreme points

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

Maximum (0, 9), Minimum (6, - 99)

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} + 9 : 9

Maximum (0, 9)

Plug

x = 6

into

x^{3} - 9 x^{2} + 9 : - 99

Minimum (6, - 99)

Maximum (0, 9), Minimum (6, - 99)

f (x) = x - 6 + 1

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

- 3 sin (4 x)

f (x) = \frac{6}{x - 5}

f (x) = \frac{12 x + 35}{x ( x + 7 )}

What is the extreme f(x)=x^3-9x^2+9 ?
The extreme f(x)=x^3-9x^2+9 is Maximum(0,9),Minimum(6,-99)