What is the extreme f(x)=x^4-2x^2+3 ?

The extreme f(x)=x^4-2x^2+3 is Minimum(-1,2),Maximum(0,3),Minimum(1,2)

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extreme f(x)=x^4-2x^2+3

extreme points

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

Minimum (- 1, 2), Maximum (0, 3), 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} + 3 : 2

Minimum (- 1, 2)

Plug

x = 0

into

x^{4} - 2 x^{2} + 3 : 3

Maximum (0, 3)

Plug

x = 1

into

x^{4} - 2 x^{2} + 3 : 2

Minimum (1, 2)

Minimum (- 1, 2), Maximum (0, 3), Minimum (1, 2)

(x - 2)^{2} - 9

f (x) = 1 - \frac{x}{9}

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

(3 - 2 x) (6 - x)

f (x) = - 2 x + 2

What is the extreme f(x)=x^4-2x^2+3 ?
The extreme f(x)=x^4-2x^2+3 is Minimum(-1,2),Maximum(0,3),Minimum(1,2)