What is the extreme f(x)=x^4-8x^2+1 ?

The extreme f(x)=x^4-8x^2+1 is Minimum(-2,-15),Maximum(0,1),Minimum(2,-15)

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extreme f(x)=x^4-8x^2+1

extreme points

f (x) = x^{4} - 8 x^{2} + 1

Minimum (- 2, - 15), Maximum (0, 1), Minimum (2, - 15)

Solution steps

f^{'} (x) = 4 x^{3} - 16 x

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

x^{4} - 8 x^{2} + 1 : - 15

Minimum (- 2, - 15)

Plug

x = 0

into

x^{4} - 8 x^{2} + 1 : 1

Maximum (0, 1)

Plug

x = 2

into

x^{4} - 8 x^{2} + 1 : - 15

Minimum (2, - 15)

Minimum (- 2, - 15), Maximum (0, 1), Minimum (2, - 15)

f (x) = x^{4} - 8 x^{2} + 6

f (x, y) = 4 x^{2} + y^{2} - 8 x + 6 y + 8

f (x) = ln (27 + 8 x^{3})

f (x) = x^{5} + 10 x^{4} + 25 x^{3}

f (x, y) = x \cdot y

What is the extreme f(x)=x^4-8x^2+1 ?
The extreme f(x)=x^4-8x^2+1 is Minimum(-2,-15),Maximum(0,1),Minimum(2,-15)