What is the extreme g(x)=x^3-3x^2+1 ?

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

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extreme g(x)=x^3-3x^2+1

extreme points

g (x) = x^{3} - 3 x^{2} + 1

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

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} + 1 : 1

Maximum (0, 1)

Plug

x = 2

into

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

Minimum (2, - 3)

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

f (x) = \frac{( x - 1 ) ^{2}}{x}

f (x, y) = 4 x y + \frac{2}{x} + \frac{1}{y}

f (x) = x^{4} - 8 x^{3} + 6

f (x) = \frac{x ^{3}}{3 - x ^{2}}

f (x) = - \frac{6 x}{3 x ^{2} + 8}

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