What is the extreme f(x)=x^4-2x^2-1 ?

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

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extreme f(x)=x^4-2x^2-1

extreme points

f (x) = x^{4} - 2 x^{2} - 1

Minimum (- 1, - 2), Maximum (0, - 1), Minimum (1, - 2)

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} - 1 : - 2

Minimum (- 1, - 2)

Plug

x = 0

into

x^{4} - 2 x^{2} - 1 : - 1

Maximum (0, - 1)

Plug

x = 1

into

x^{4} - 2 x^{2} - 1 : - 2

Minimum (1, - 2)

Minimum (- 1, - 2), Maximum (0, - 1), Minimum (1, - 2)

f (x, y) = x^{3} - 6 x y + y^{3}

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

f (x) = x ln (2 x)

f (x, y) = x^{2} - x + 2 y^{2}

f (x) = (x - 1)^{2}

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