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

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

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extreme f(x)=x^4-8x^2

extreme points

f (x) = 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)

f (x) = x^{2} - 6 x + 1

f (x) = x^{2} - x - 20

f (x) = cos (4 x)

9 x - y = - 18, (0, 0)

f (x) = - x - 1 - 2

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