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

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

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

extreme points

f (x) = 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)

y = csc (x)

f (x) = \frac{x + 7}{x ^{2} - 49}

f (x) = x^{2} + 9 x + 2

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

f (x) = \frac{60}{x ( x + 4 )}

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