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

The extreme f(x)=x^3-3x^2+3 is Maximum(0,3),Minimum(2,-1)

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

extreme points

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

Maximum (0, 3), Minimum (2, - 1)

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} + 3 : 3

Maximum (0, 3)

Plug

x = 2

into

x^{3} - 3 x^{2} + 3 : - 1

Minimum (2, - 1)

Maximum (0, 3), Minimum (2, - 1)

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

f (x, y) = \frac{1}{x ^{2} + y ^{2} + 1}

f (x, y) = x y

f (x) = x 8 - x^{2}

f (x, y) = 4 x y - x^{4} - 2 y^{2} + 2

What is the extreme f(x)=x^3-3x^2+3 ?
The extreme f(x)=x^3-3x^2+3 is Maximum(0,3),Minimum(2,-1)