What is the extreme f(x)=2x^4+8x^2-8x^3-1 ?

The extreme f(x)=2x^4+8x^2-8x^3-1 is Minimum(0,-1),Maximum(1,1),Minimum(2,-1)

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extreme f(x)=2x^4+8x^2-8x^3-1

extreme

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

Solution steps

f^{'} (x) = 8 x^{3} - 24 x^{2} + 16 x

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

2 x^{4} + 8 x^{2} - 8 x^{3} - 1 : - 1

Minimum (0, - 1)

Plug

x = 1

into

2 x^{4} + 8 x^{2} - 8 x^{3} - 1 : 1

Maximum (1, 1)

Plug

x = 2

into

2 x^{4} + 8 x^{2} - 8 x^{3} - 1 : - 1

Minimum (2, - 1)

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

What is the extreme f(x)=2x^4+8x^2-8x^3-1 ?
The extreme f(x)=2x^4+8x^2-8x^3-1 is Minimum(0,-1),Maximum(1,1),Minimum(2,-1)