What is the extreme y=x^4-2x^2+1 ?

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

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extreme y=x^4-2x^2+1

extreme points

y = x^{4} - 2 x^{2} + 1

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

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 : 0

Minimum (- 1, 0)

Plug

x = 0

into

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

Maximum (0, 1)

Plug

x = 1

into

x^{4} - 2 x^{2} + 1 : 0

Minimum (1, 0)

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

f (x) = 8 sin (2 x)

f (x) = - 20 x^{2} - 30 y^{2} + 560 x + 360 y + 180

\frac{3 x + 5}{5 x ^{2} + 55 x + 120}

f (x) = x^{4} e^{- 7 x}

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