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

The extreme f(x)=x^4-8x^2+4 is Minimum(-2,-12),Maximum(0,4),Minimum(2,-12)

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

extreme points

f (x) = x^{4} - 8 x^{2} + 4

Minimum (- 2, - 12), Maximum (0, 4), Minimum (2, - 12)

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} + 4 : - 12

Minimum (- 2, - 12)

Plug

x = 0

into

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

Maximum (0, 4)

Plug

x = 2

into

x^{4} - 8 x^{2} + 4 : - 12

Minimum (2, - 12)

Minimum (- 2, - 12), Maximum (0, 4), Minimum (2, - 12)

5 x - 10 y = 20

f (x) = 7 x - 5

f (x) = arcsin (ln (x))

y = x^{3} + 12 x^{2} + 48 x - 1

f (x) = \frac{x}{8 x + 7}

What is the extreme f(x)=x^4-8x^2+4 ?
The extreme f(x)=x^4-8x^2+4 is Minimum(-2,-12),Maximum(0,4),Minimum(2,-12)