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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = 0

into

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

Minimum (0, 1)

Plug

x = 1

into

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

Maximum (1, 2)

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

f (x) = \frac{1}{4} x + 3 + \frac{400}{x}

f (x) = cos (x) - x

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

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

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

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