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

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

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

extreme points

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

Minimum (- 2, 0), Maximum (0, 4), Minimum (2, 0)

Solution steps

f^{'} (x) = 4 x^{3} - 8 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} - 4 x^{2} + 4 : 0

Minimum (- 2, 0)

Plug

x = 0

into

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

Maximum (0, 4)

Plug

x = 2

into

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

Minimum (2, 0)

Minimum (- 2, 0), Maximum (0, 4), Minimum (2, 0)

f (x) = - 3 x - 2

2 x - 3 y = - 6

1 + ln (t)

g (x) = x^{2} - 4 x - 21

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