What is the extreme y=x^4-8x^2 ?

The extreme y=x^4-8x^2 is Minimum(-2,-16),Maximum(0,0),Minimum(2,-16)

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extreme y=x^4-8x^2

extreme points

y = x^{4} - 8 x^{2}

Minimum (- 2, - 16), Maximum (0, 0), Minimum (2, - 16)

Solution steps

f^{'} (x) = 4 x^{3} - 16 x

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

x^{4} - 8 x^{2} : - 16

Minimum (- 2, - 16)

Plug

x = 0

into

x^{4} - 8 x^{2} : 0

Maximum (0, 0)

Plug

x = 2

into

x^{4} - 8 x^{2} : - 16

Minimum (2, - 16)

Minimum (- 2, - 16), Maximum (0, 0), Minimum (2, - 16)

What is the extreme y=x^4-8x^2 ?
The extreme y=x^4-8x^2 is Minimum(-2,-16),Maximum(0,0),Minimum(2,-16)