What is the extreme f(x)=x^4-8x^2+18 ?

The extreme f(x)=x^4-8x^2+18 is Minimum(-2,2),Maximum(0,18),Minimum(2,2)

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extreme f(x)=x^4-8x^2+18

extreme points

f (x) = x^{4} - 8 x^{2} + 18

Minimum (- 2, 2), Maximum (0, 18), Minimum (2, 2)

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} + 18 : 2

Minimum (- 2, 2)

Plug

x = 0

into

x^{4} - 8 x^{2} + 18 : 18

Maximum (0, 18)

Plug

x = 2

into

x^{4} - 8 x^{2} + 18 : 2

Minimum (2, 2)

Minimum (- 2, 2), Maximum (0, 18), Minimum (2, 2)

f (x) = x^{3} + 29 x + 128

x^{2} + \frac{4}{x}

f (x) = x e^{- x} [0.2]

f (x) = - 2 x^{2} - 12 x - 23

x^{2} - x y + y^{2} + 3 x - 2 y - 5

What is the extreme f(x)=x^4-8x^2+18 ?
The extreme f(x)=x^4-8x^2+18 is Minimum(-2,2),Maximum(0,18),Minimum(2,2)