What is the extreme f(x)=2x^4-196x^2-2 ?

The extreme f(x)=2x^4-196x^2-2 is Minimum(-7,-4804),Maximum(0,-2),Minimum(7,-4804)

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extreme f(x)=2x^4-196x^2-2

extreme points

f (x) = 2 x^{4} - 196 x^{2} - 2

Minimum (- 7, - 4804), Maximum (0, - 2), Minimum (7, - 4804)

Solution steps

f^{'} (x) = 8 x^{3} - 392 x

Find intervals:

Decreasing

: - \infty < x < - 7,

Increasing

: - 7 < x < 0,

Decreasing

: 0 < x < 7,

Increasing

: 7 < x < \infty

Plug

x = - 7

into

2 x^{4} - 196 x^{2} - 2 : - 4804

Minimum (- 7, - 4804)

Plug

x = 0

into

2 x^{4} - 196 x^{2} - 2 : - 2

Maximum (0, - 2)

Plug

x = 7

into

2 x^{4} - 196 x^{2} - 2 : - 4804

Minimum (7, - 4804)

Minimum (- 7, - 4804), Maximum (0, - 2), Minimum (7, - 4804)

f (x) = 2 x^{4} - 196 x^{2} + 4

f (x) = ∣ 3 x - 4 ∣

f (x, y) = 8 - 3 x^{2} + 6 x - 2 y^{2} + 8 y

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

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

What is the extreme f(x)=2x^4-196x^2-2 ?
The extreme f(x)=2x^4-196x^2-2 is Minimum(-7,-4804),Maximum(0,-2),Minimum(7,-4804)