What is the extreme f(x)=3x^4-96x^2+1 ?

The extreme f(x)=3x^4-96x^2+1 is Minimum(-4,-767),Maximum(0,1),Minimum(4,-767)

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extreme f(x)=3x^4-96x^2+1

extreme points

f (x) = 3 x^{4} - 96 x^{2} + 1

Minimum (- 4, - 767), Maximum (0, 1), Minimum (4, - 767)

Solution steps

f^{'} (x) = 12 x^{3} - 192 x

Find intervals:

Decreasing

: - \infty < x < - 4,

Increasing

: - 4 < x < 0,

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = - 4

into

3 x^{4} - 96 x^{2} + 1 : - 767

Minimum (- 4, - 767)

Plug

x = 0

into

3 x^{4} - 96 x^{2} + 1 : 1

Maximum (0, 1)

Plug

x = 4

into

3 x^{4} - 96 x^{2} + 1 : - 767

Minimum (4, - 767)

Minimum (- 4, - 767), Maximum (0, 1), Minimum (4, - 767)

f (x) = 0.003 x^{2} + 4.2 x - 50

f (x) = - 3 x^{2} + 1

f (x) = - 3 x^{2} + 3

f (x) = 2 x^{3} + x^{2} - 20 x

f (x) = 2 x^{2} - 4 x [0.3]

What is the extreme f(x)=3x^4-96x^2+1 ?
The extreme f(x)=3x^4-96x^2+1 is Minimum(-4,-767),Maximum(0,1),Minimum(4,-767)