What is the extreme f(x)=x^4-162x^2+6561 ?

The extreme f(x)=x^4-162x^2+6561 is Minimum(-9,0),Maximum(0,6561),Minimum(9,0)

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extreme f(x)=x^4-162x^2+6561

extreme points

f (x) = x^{4} - 162 x^{2} + 6561

Minimum (- 9, 0), Maximum (0, 6561), Minimum (9, 0)

Solution steps

f^{'} (x) = 4 x^{3} - 324 x

Find intervals:

Decreasing

: - \infty < x < - 9,

Increasing

: - 9 < x < 0,

Decreasing

: 0 < x < 9,

Increasing

: 9 < x < \infty

Plug

x = - 9

into

x^{4} - 162 x^{2} + 6561 : 0

Minimum (- 9, 0)

Plug

x = 0

into

x^{4} - 162 x^{2} + 6561 : 6561

Maximum (0, 6561)

Plug

x = 9

into

x^{4} - 162 x^{2} + 6561 : 0

Minimum (9, 0)

Minimum (- 9, 0), Maximum (0, 6561), Minimum (9, 0)

f (x) = - 3 x + 5

y = - \frac{2}{3} x + 4

f (x) = x (x + 4)

\frac{1}{2} 3 x

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

What is the extreme f(x)=x^4-162x^2+6561 ?
The extreme f(x)=x^4-162x^2+6561 is Minimum(-9,0),Maximum(0,6561),Minimum(9,0)