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

The extreme f(x)=x^4-8x^2+2 is Minimum(-2,-14),Maximum(0,2),Minimum(2,-14)

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

extreme points

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

Minimum (- 2, - 14), Maximum (0, 2), Minimum (2, - 14)

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} + 2 : - 14

Minimum (- 2, - 14)

Plug

x = 0

into

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

Maximum (0, 2)

Plug

x = 2

into

x^{4} - 8 x^{2} + 2 : - 14

Minimum (2, - 14)

Minimum (- 2, - 14), Maximum (0, 2), Minimum (2, - 14)

f (x) = lo g_{\frac{1}{5}} (x)

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

f (x) = ln (x + 3)

y = - x^{3} + 3

What is the extreme f(x)=x^4-8x^2+2 ?
The extreme f(x)=x^4-8x^2+2 is Minimum(-2,-14),Maximum(0,2),Minimum(2,-14)