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

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

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

extreme points

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

Maximum (0, 1), Minimum (1, 0)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = 0

into

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

Maximum (0, 1)

Plug

x = 1

into

2 x^{3} - 3 x^{2} + 1 : 0

Minimum (1, 0)

Maximum (0, 1), Minimum (1, 0)

f (x, y) = 20 x + 64 y + 8 x y - x^{2} - 32 y^{2}

y = \frac{x}{1 + x ^{2}}

\frac{x ^{2} - 1}{x ^{2} - 4}

f (x) = 5 x^{2} ln (x)

f (x, y) = 2 x^{2} + 8 x y + y^{4}

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