What is the extreme f(x)=x^3-3x^2+6 ?

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

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extreme f(x)=x^3-3x^2+6

extreme points

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

Maximum (0, 6), Minimum (2, 2)

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} + 6 : 6

Maximum (0, 6)

Plug

x = 2

into

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

Minimum (2, 2)

Maximum (0, 6), Minimum (2, 2)

f (x, y) = x^{2} + y^{2} - 2 x - 6 y + 14

f (x, y) = x^{3} - y^{3} - 2 x y + 6

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

f (t) = 2 x

f (x, y) = 12 x y^{2} - 16 x + 25 x y^{3} - 18 x^{2} y^{3}

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