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

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

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

extreme points

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

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

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} + 2 : 1

Minimum (- 1, 1)

Plug

x = 0

into

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

Maximum (0, 2)

Plug

x = 1

into

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

Minimum (1, 1)

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

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

f (x) = 4 x^{3} + 3 x^{2} - 6 x

f (x) = x^{5} - 5 x

f (x, y) = \frac{x y}{x - 2 y}

f (x) = - 3 x^{\frac{5}{3}} - 15 x^{\frac{2}{3}}

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