What is the extreme f(x)=x^3-3x^2+12 ?

The extreme f(x)=x^3-3x^2+12 is Maximum(0,12),Minimum(2,8)

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extreme f(x)=x^3-3x^2+12

extreme points

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

Maximum (0, 12), Minimum (2, 8)

Solution steps

f^{'} (x) = 3 x^{2} - 6 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

x^{3} - 3 x^{2} + 12 : 12

Maximum (0, 12)

Plug

x = 2

into

x^{3} - 3 x^{2} + 12 : 8

Minimum (2, 8)

Maximum (0, 12), Minimum (2, 8)

What is the extreme f(x)=x^3-3x^2+12 ?
The extreme f(x)=x^3-3x^2+12 is Maximum(0,12),Minimum(2,8)