What is the extreme y=x^4-8x^2+3 ?

The extreme y=x^4-8x^2+3 is Minimum(-2,-13),Maximum(0,3),Minimum(2,-13)

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extreme y=x^4-8x^2+3

extreme points

y = x^{4} - 8 x^{2} + 3

Minimum (- 2, - 13), Maximum (0, 3), Minimum (2, - 13)

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} + 3 : - 13

Minimum (- 2, - 13)

Plug

x = 0

into

x^{4} - 8 x^{2} + 3 : 3

Maximum (0, 3)

Plug

x = 2

into

x^{4} - 8 x^{2} + 3 : - 13

Minimum (2, - 13)

Minimum (- 2, - 13), Maximum (0, 3), Minimum (2, - 13)

f (x) = 2 x^{3} - 5 x^{2} + 3

f (x, y) = 2 x^{2} + 3 y^{2} + 2 x y + 14 x - 28 y

f (x) = x^{2} - 6 x + 14

f (x, y) = 10 x^{2} y - 5 x^{2} - 4 y^{2} - x^{4} - 2 y^{4}

f (x) = x^{\frac{4}{5}} (x - 9)^{2}

What is the extreme y=x^4-8x^2+3 ?
The extreme y=x^4-8x^2+3 is Minimum(-2,-13),Maximum(0,3),Minimum(2,-13)