What is the extreme f(x)=x^4-32x^2-4 ?

The extreme f(x)=x^4-32x^2-4 is Minimum(-4,-260),Maximum(0,-4),Minimum(4,-260)

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extreme f(x)=x^4-32x^2-4

extreme points

f (x) = x^{4} - 32 x^{2} - 4

Minimum (- 4, - 260), Maximum (0, - 4), Minimum (4, - 260)

Solution steps

f^{'} (x) = 4 x^{3} - 64 x

Find intervals:

Decreasing

: - \infty < x < - 4,

Increasing

: - 4 < x < 0,

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = - 4

into

x^{4} - 32 x^{2} - 4 : - 260

Minimum (- 4, - 260)

Plug

x = 0

into

x^{4} - 32 x^{2} - 4 : - 4

Maximum (0, - 4)

Plug

x = 4

into

x^{4} - 32 x^{2} - 4 : - 260

Minimum (4, - 260)

Minimum (- 4, - 260), Maximum (0, - 4), Minimum (4, - 260)

What is the extreme f(x)=x^4-32x^2-4 ?
The extreme f(x)=x^4-32x^2-4 is Minimum(-4,-260),Maximum(0,-4),Minimum(4,-260)