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

The extreme f(x)=2x^4-196x^2+4 is Minimum(-7,-4798),Maximum(0,4),Minimum(7,-4798)

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

extreme

Minimum (- 7, - 4798), Maximum (0, 4), Minimum (7, - 4798)

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} + 4 : - 4798

Minimum (- 7, - 4798)

Plug

x = 0

into

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

Maximum (0, 4)

Plug

x = 7

into

2 x^{4} - 196 x^{2} + 4 : - 4798

Minimum (7, - 4798)

Minimum (- 7, - 4798), Maximum (0, 4), Minimum (7, - 4798)

What is the extreme f(x)=2x^4-196x^2+4 ?
The extreme f(x)=2x^4-196x^2+4 is Minimum(-7,-4798),Maximum(0,4),Minimum(7,-4798)