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

The extreme f(x)=(x^2-4)^2-4 is Minimum(-2,-4),Maximum(0,12),Minimum(2,-4)

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

extreme points

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

Minimum (- 2, - 4), Maximum (0, 12), Minimum (2, - 4)

Solution steps

f^{'} (x) = 4 x^{3} - 16 x

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} - 4 : - 4

Minimum (- 2, - 4)

Plug

x = 0

into

(x^{2} - 4)^{2} - 4 : 12

Maximum (0, 12)

Plug

x = 2

into

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

Minimum (2, - 4)

Minimum (- 2, - 4), Maximum (0, 12), Minimum (2, - 4)

f (x) = \frac{2}{3} x^{3} - 4 x^{2} - 45 x

f (x, y) = ln (x + y + 4)

2 x^{3} - 3 x^{2} - 12 x + 5

f (x) = x^{2} - x y + y^{2} + 6 y + 2

\frac{x}{x ^{2} + 10 x + 21}

What is the extreme f(x)=(x^2-4)^2-4 ?
The extreme f(x)=(x^2-4)^2-4 is Minimum(-2,-4),Maximum(0,12),Minimum(2,-4)