What is the extreme y=(x^2-9)^3 ?

The extreme y=(x^2-9)^3 is Saddle(-3,0),Minimum(0,-729),Saddle(3,0)

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extreme y=(x^2-9)^3

extreme points

y = (x^{2} - 9)^{3}

Saddle (- 3, 0), Minimum (0, - 729), Saddle (3, 0)

Solution steps

f^{'} (x) = 6 x (x^{2} - 9)^{2}

Find intervals:

Decreasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < 0,

Increasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

(x^{2} - 9)^{3} : 0

Saddle (- 3, 0)

Plug

x = 0

into

(x^{2} - 9)^{3} : - 729

Minimum (0, - 729)

Plug

x = 3

into

(x^{2} - 9)^{3} : 0

Saddle (3, 0)

Saddle (- 3, 0), Minimum (0, - 729), Saddle (3, 0)

f (x) = \frac{1}{x}, [1, \infty]

f (x) = - 0.1 x^{2} + 1.4 x + 98.9

f (x, y) = 6 y^{2} + 4 x^{2} + 4 x y - 16 y + 8 x

y = (lo g_{10} (5 x)) + (3 I n x^{3})

x^{2} + \frac{640}{x}

What is the extreme y=(x^2-9)^3 ?
The extreme y=(x^2-9)^3 is Saddle(-3,0),Minimum(0,-729),Saddle(3,0)