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

The extreme f(x)=(x^2-4)^2 is Minimum(-2,0),Maximum(0,16),Minimum(2,0)

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

extreme points

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

Minimum (- 2, 0), Maximum (0, 16), Minimum (2, 0)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

(x^{2} - 4)^{2} : 0

Minimum (- 2, 0)

Plug

x = 0

into

(x^{2} - 4)^{2} : 16

Maximum (0, 16)

Plug

x = 2

into

(x^{2} - 4)^{2} : 0

Minimum (2, 0)

Minimum (- 2, 0), Maximum (0, 16), Minimum (2, 0)

p er p e n d i c u l a r (1, 6), y = - \frac{1}{4} x + 3

in f l ec t i o n f (x) = x^{\frac{2}{5}} (x - 5)

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

cr i t i c a l - x^{2} + 8 x - 9

p er p e n d i c u l a r x + 2 y = 16

What is the extreme f(x)=(x^2-4)^2 ?
The extreme f(x)=(x^2-4)^2 is Minimum(-2,0),Maximum(0,16),Minimum(2,0)