What is the extreme y=(x^2-9)^4 ?

The extreme y=(x^2-9)^4 is Minimum(-3,0),Maximum(0,6561),Minimum(3,0)

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extreme y=(x^2-9)^4

extreme points

y = (x^{2} - 9)^{4}

Minimum (- 3, 0), Maximum (0, 6561), Minimum (3, 0)

Solution steps

f^{'} (x) = 8 x (x^{2} - 9)^{3}

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

(x^{2} - 9)^{4} : 0

Minimum (- 3, 0)

Plug

x = 0

into

(x^{2} - 9)^{4} : 6561

Maximum (0, 6561)

Plug

x = 3

into

(x^{2} - 9)^{4} : 0

Minimum (3, 0)

Minimum (- 3, 0), Maximum (0, 6561), Minimum (3, 0)

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

f (x, y) = e^{y} - y e^{x}

f (x) = 3 x^{2} + 5 x - 4

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

f (x, y) = x^{2} + x y - (7 + 1) y - (8 + 1) x + 1

What is the extreme y=(x^2-9)^4 ?
The extreme y=(x^2-9)^4 is Minimum(-3,0),Maximum(0,6561),Minimum(3,0)