What is the extreme f(x)=(x^2-4)^3 ?

The extreme f(x)=(x^2-4)^3 is Saddle(-2,0),Minimum(0,-64),Saddle(2,0)

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extreme f(x)=(x^2-4)^3

extreme points

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

Saddle (- 2, 0), Minimum (0, - 64), Saddle (2, 0)

Solution steps

f^{'} (x) = 6 x (x^{2} - 4)^{2}

Find intervals:

Decreasing

: - \infty < x < - 2,

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

(x^{2} - 4)^{3} : 0

Saddle (- 2, 0)

Plug

x = 0

into

(x^{2} - 4)^{3} : - 64

Minimum (0, - 64)

Plug

x = 2

into

(x^{2} - 4)^{3} : 0

Saddle (2, 0)

Saddle (- 2, 0), Minimum (0, - 64), Saddle (2, 0)

f (x, y) = 6 - \frac{x ^{2}}{3} - \frac{y ^{2}}{2}

f (x) = x^{8} ln (x)

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

f (x) = ln (x^{2} + 2)

- lo g_{5} (x)

What is the extreme f(x)=(x^2-4)^3 ?
The extreme f(x)=(x^2-4)^3 is Saddle(-2,0),Minimum(0,-64),Saddle(2,0)