What is the extreme y=2x^4-12x^2 ?

The extreme y=2x^4-12x^2 is Minimum(-sqrt(3),-18),Maximum(0,0),Minimum(sqrt(3),-18)

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extreme y=2x^4-12x^2

extreme points

y = 2 x^{4} - 12 x^{2}

Minimum (- 3, - 18), Maximum (0, 0), Minimum (3, - 18)

Solution steps

f^{'} (x) = 8 x^{3} - 24 x

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

2 x^{4} - 12 x^{2} : - 18

Minimum (- 3, - 18)

Plug

x = 0

into

2 x^{4} - 12 x^{2} : 0

Maximum (0, 0)

Plug

x = 3

into

2 x^{4} - 12 x^{2} : - 18

Minimum (3, - 18)

Minimum (- 3, - 18), Maximum (0, 0), Minimum (3, - 18)

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

- x^{2} - x + 1

\frac{x ^{2} + x}{∣ x ∣ + 1}

18 x y - 3 x^{2} y - 2 x y^{2}

f (x) = x^{2} - 8 x - 9, 1 \leq x \leq 6

What is the extreme y=2x^4-12x^2 ?
The extreme y=2x^4-12x^2 is Minimum(-sqrt(3),-18),Maximum(0,0),Minimum(sqrt(3),-18)