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

The extreme f(x)=x^3-9x^2+3 is Maximum(0,3),Minimum(6,-105)

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

extreme points

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

Maximum (0, 3), Minimum (6, - 105)

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} + 3 : 3

Maximum (0, 3)

Plug

x = 6

into

x^{3} - 9 x^{2} + 3 : - 105

Minimum (6, - 105)

Maximum (0, 3), Minimum (6, - 105)

f (x) = lo g_{b} (x)

y = 9 x^{2} - 4

y = 3 x^{2} + 6 x - 12

f (x) = x^{4} + 16 x^{2} + 72

y = 3 x + 2

What is the extreme f(x)=x^3-9x^2+3 ?
The extreme f(x)=x^3-9x^2+3 is Maximum(0,3),Minimum(6,-105)