What is the extreme f(x)=18x^4-108x^2 ?

The extreme f(x)=18x^4-108x^2 is Minimum(-sqrt(3),-162),Maximum(0,0),Minimum(sqrt(3),-162)

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extreme f(x)=18x^4-108x^2

extreme points

f (x) = 18 x^{4} - 108 x^{2}

Minimum (- 3, - 162), Maximum (0, 0), Minimum (3, - 162)

Solution steps

f^{'} (x) = 72 x^{3} - 216 x

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

18 x^{4} - 108 x^{2} : - 162

Minimum (- 3, - 162)

Plug

x = 0

into

18 x^{4} - 108 x^{2} : 0

Maximum (0, 0)

Plug

x = 3

into

18 x^{4} - 108 x^{2} : - 162

Minimum (3, - 162)

Minimum (- 3, - 162), Maximum (0, 0), Minimum (3, - 162)

f (x) = x^{2} + \frac{160}{x}

f (x) = sin^{2} (x)

f (x) = 4 x

(7, 2), (7, 7)

f (x) = \frac{2}{3} x + 1

What is the extreme f(x)=18x^4-108x^2 ?
The extreme f(x)=18x^4-108x^2 is Minimum(-sqrt(3),-162),Maximum(0,0),Minimum(sqrt(3),-162)