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

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

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

extreme points

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

Minimum (0, 4), Maximum (1, 6)

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

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

Minimum (0, 4)

Plug

x = 1

into

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

Maximum (1, 6)

Minimum (0, 4), Maximum (1, 6)

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