What is the extreme f(x)=4-12x^2+1/16 x^4 ?

The extreme f(x)=4-12x^2+1/16 x^4 is Minimum(-4sqrt(6),-572),Maximum(0,4),Minimum(4sqrt(6),-572)

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extreme f(x)=4-12x^2+1/16 x^4

extreme points

f (x) = 4 - 12 x^{2} + \frac{1}{16} x^{4}

Minimum (- 46, - 572), Maximum (0, 4), Minimum (46, - 572)

Solution steps

f^{'} (x) = \frac{x ^{3}}{4} - 24 x

Find intervals:

Decreasing

: - \infty < x < - 46,

Increasing

: - 46 < x < 0,

Decreasing

: 0 < x < 46,

Increasing

: 46 < x < \infty

Plug

x = - 46

into

4 - 12 x^{2} + \frac{1}{16} x^{4} : - 572

Minimum (- 46, - 572)

Plug

x = 0

into

4 - 12 x^{2} + \frac{1}{16} x^{4} : 4

Maximum (0, 4)

Plug

x = 46

into

4 - 12 x^{2} + \frac{1}{16} x^{4} : - 572

Minimum (46, - 572)

Minimum (- 46, - 572), Maximum (0, 4), Minimum (46, - 572)

y = \frac{( 3 x - 4 )}{- 2 x + 2}

y = x^{2} - 2

f (x) = 9 x + 3

f (x) = ln (x - 2)

What is the extreme f(x)=4-12x^2+1/16 x^4 ?
The extreme f(x)=4-12x^2+1/16 x^4 is Minimum(-4sqrt(6),-572),Maximum(0,4),Minimum(4sqrt(6),-572)