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

The extreme f(x)=x^4-18x^2 is Minimum(-3,-81),Maximum(0,0),Minimum(3,-81)

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

extreme points

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

Minimum (- 3, - 81), Maximum (0, 0), Minimum (3, - 81)

Solution steps

f^{'} (x) = 4 x^{3} - 36 x

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

x^{4} - 18 x^{2} : - 81

Minimum (- 3, - 81)

Plug

x = 0

into

x^{4} - 18 x^{2} : 0

Maximum (0, 0)

Plug

x = 3

into

x^{4} - 18 x^{2} : - 81

Minimum (3, - 81)

Minimum (- 3, - 81), Maximum (0, 0), Minimum (3, - 81)

f (x) = x^{3} - 3 x^{2} - 13 x + 15

f (x) = x^{\frac{5}{2}} - 8 x^{2}

(1, 13), (27, 21)

f (x) = ln (x^{2} + 1)

(3.2) (5.4)

What is the extreme f(x)=x^4-18x^2 ?
The extreme f(x)=x^4-18x^2 is Minimum(-3,-81),Maximum(0,0),Minimum(3,-81)