What is the extreme 2x^4-196x^2-3 ?

The extreme 2x^4-196x^2-3 is Minimum(-7,-4805),Maximum(0,-3),Minimum(7,-4805)

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extreme 2x^4-196x^2-3

extreme

Minimum (- 7, - 4805), Maximum (0, - 3), Minimum (7, - 4805)

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} - 3 : - 4805

Minimum (- 7, - 4805)

Plug

x = 0

into

2 x^{4} - 196 x^{2} - 3 : - 3

Maximum (0, - 3)

Plug

x = 7

into

2 x^{4} - 196 x^{2} - 3 : - 4805

Minimum (7, - 4805)

Minimum (- 7, - 4805), Maximum (0, - 3), Minimum (7, - 4805)

What is the extreme 2x^4-196x^2-3 ?
The extreme 2x^4-196x^2-3 is Minimum(-7,-4805),Maximum(0,-3),Minimum(7,-4805)