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

The extreme f(x)=x^4-2x^2+8 is Minimum(-1,7),Maximum(0,8),Minimum(1,7)

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

extreme points

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

Minimum (- 1, 7), Maximum (0, 8), Minimum (1, 7)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 1

into

x^{4} - 2 x^{2} + 8 : 7

Minimum (- 1, 7)

Plug

x = 0

into

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

Maximum (0, 8)

Plug

x = 1

into

x^{4} - 2 x^{2} + 8 : 7

Minimum (1, 7)

Minimum (- 1, 7), Maximum (0, 8), Minimum (1, 7)

f (x) = - \frac{1}{x - 2}

f (x) = (8 - (3 x^{2} + 2 x + 1))

f (x) = \frac{( 2 x ^{2} )}{9 - x ^{2}}

f (x) = - \frac{1}{x - 6}

f (x, y) = 18 x^{2} - 3 y^{2} - 36 x - 128 y - 110

What is the extreme f(x)=x^4-2x^2+8 ?
The extreme f(x)=x^4-2x^2+8 is Minimum(-1,7),Maximum(0,8),Minimum(1,7)