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

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

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

extreme points

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

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

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

Minimum (- 4, - 256)

Plug

x = 0

into

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

Maximum (0, 0)

Plug

x = 4

into

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

Minimum (4, - 256)

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

f (x) = \frac{x + 3}{x ( x + 11 )}

f (x) = \frac{1}{5 x - 2}

f (x) = - 5 x^{3} + 15 x + 3

y = \frac{4}{x}

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

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