What is the extreme f(x)=3+6x^2-4x^3 ?

The extreme f(x)=3+6x^2-4x^3 is Minimum(0,3),Maximum(1,5)

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extreme f(x)=3+6x^2-4x^3

extreme points

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

Minimum (0, 3), Maximum (1, 5)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = 0

into

3 + 6 x^{2} - 4 x^{3} : 3

Minimum (0, 3)

Plug

x = 1

into

3 + 6 x^{2} - 4 x^{3} : 5

Maximum (1, 5)

Minimum (0, 3), Maximum (1, 5)

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

f (x) = - 5 x^{3} + 60 x + 2

f (x) = - 5 x^{3} + 60 x + 3

f (x) = x^{3} - 5 x^{2} - 8 x + 2

f (x, y) = 9 x^{2} + x y + 10 y^{2}

What is the extreme f(x)=3+6x^2-4x^3 ?
The extreme f(x)=3+6x^2-4x^3 is Minimum(0,3),Maximum(1,5)