What is the extreme f(x)=x^4-128x^2+4096 ?

The extreme f(x)=x^4-128x^2+4096 is Minimum(-8,0),Maximum(0,4096),Minimum(8,0)

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extreme f(x)=x^4-128x^2+4096

extreme points

f (x) = x^{4} - 128 x^{2} + 4096

Minimum (- 8, 0), Maximum (0, 4096), Minimum (8, 0)

Solution steps

f^{'} (x) = 4 x^{3} - 256 x

Find intervals:

Decreasing

: - \infty < x < - 8,

Increasing

: - 8 < x < 0,

Decreasing

: 0 < x < 8,

Increasing

: 8 < x < \infty

Plug

x = - 8

into

x^{4} - 128 x^{2} + 4096 : 0

Minimum (- 8, 0)

Plug

x = 0

into

x^{4} - 128 x^{2} + 4096 : 4096

Maximum (0, 4096)

Plug

x = 8

into

x^{4} - 128 x^{2} + 4096 : 0

Minimum (8, 0)

Minimum (- 8, 0), Maximum (0, 4096), Minimum (8, 0)

f (x) = x^{2 x}

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

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

y = \frac{x}{x + 1}

y = 2 (x + 3) (x - 7)

What is the extreme f(x)=x^4-128x^2+4096 ?
The extreme f(x)=x^4-128x^2+4096 is Minimum(-8,0),Maximum(0,4096),Minimum(8,0)