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

The extreme f(x)=x^4-2x^2-8 is Minimum(-1,-9),Maximum(0,-8),Minimum(1,-9)

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

extreme points

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

Minimum (- 1, - 9), Maximum (0, - 8), Minimum (1, - 9)

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 : - 9

Minimum (- 1, - 9)

Plug

x = 0

into

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

Maximum (0, - 8)

Plug

x = 1

into

x^{4} - 2 x^{2} - 8 : - 9

Minimum (1, - 9)

Minimum (- 1, - 9), Maximum (0, - 8), Minimum (1, - 9)

f (x, y) = 2 - 1 + x + y

f (x) = 3 - 4 sin (3 x), (0, 2 π)

f (x) = \frac{2 x}{x ^{2} + 4}

\frac{x}{x ^{2} + 9}

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

What is the extreme f(x)=x^4-2x^2-8 ?
The extreme f(x)=x^4-2x^2-8 is Minimum(-1,-9),Maximum(0,-8),Minimum(1,-9)