What is the extreme f(x)=3x^3-6x^2+1 ?

The extreme f(x)=3x^3-6x^2+1 is Maximum(0,1),Minimum(4/3 ,-23/9)

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extreme f(x)=3x^3-6x^2+1

extreme points

f (x) = 3 x^{3} - 6 x^{2} + 1

Maximum (0, 1), Minimum (\frac{4}{3}, - \frac{23}{9})

Solution steps

f^{'} (x) = 9 x^{2} - 12 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{4}{3},

Increasing

: \frac{4}{3} < x < \infty

Plug

x = 0

into

3 x^{3} - 6 x^{2} + 1 : 1

Maximum (0, 1)

Plug

x = \frac{4}{3}

into

3 x^{3} - 6 x^{2} + 1 : - \frac{23}{9}

Minimum (\frac{4}{3}, - \frac{23}{9})

Maximum (0, 1), Minimum (\frac{4}{3}, - \frac{23}{9})

f (x) = x^{4} - x^{3} - 6 x^{2}

f (x) = \frac{1}{x ^{2} + 2 x + 9}

f (x) = (x - 3)^{4}

f (x, y) = 1 - x - y

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

What is the extreme f(x)=3x^3-6x^2+1 ?
The extreme f(x)=3x^3-6x^2+1 is Maximum(0,1),Minimum(4/3 ,-23/9)