What is the extreme f(x)=x^4-32x^2+256 ?

The extreme f(x)=x^4-32x^2+256 is Minimum(-4,0),Maximum(0,256),Minimum(4,0)

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extreme f(x)=x^4-32x^2+256

extreme points

f (x) = x^{4} - 32 x^{2} + 256

Minimum (- 4, 0), Maximum (0, 256), Minimum (4, 0)

Solution steps

f^{'} (x) = 4 x^{3} - 64 x

Find intervals:

Decreasing

: - \infty < x < - 4,

Increasing

: - 4 < x < 0,

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = - 4

into

x^{4} - 32 x^{2} + 256 : 0

Minimum (- 4, 0)

Plug

x = 0

into

x^{4} - 32 x^{2} + 256 : 256

Maximum (0, 256)

Plug

x = 4

into

x^{4} - 32 x^{2} + 256 : 0

Minimum (4, 0)

Minimum (- 4, 0), Maximum (0, 256), Minimum (4, 0)

x^{2} ln (\frac{x}{6})

y + 4 = - \frac{5}{2} (x - 2)

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

3 x - 2 y = 4

\frac{4}{x - 2}

What is the extreme f(x)=x^4-32x^2+256 ?
The extreme f(x)=x^4-32x^2+256 is Minimum(-4,0),Maximum(0,256),Minimum(4,0)