What is the extreme f(x)=x^4-18x^2+81 ?

The extreme f(x)=x^4-18x^2+81 is Minimum(-3,0),Maximum(0,81),Minimum(3,0)

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extreme f(x)=x^4-18x^2+81

extreme points

f (x) = x^{4} - 18 x^{2} + 81

Minimum (- 3, 0), Maximum (0, 81), Minimum (3, 0)

Solution steps

f^{'} (x) = 4 x^{3} - 36 x

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

x^{4} - 18 x^{2} + 81 : 0

Minimum (- 3, 0)

Plug

x = 0

into

x^{4} - 18 x^{2} + 81 : 81

Maximum (0, 81)

Plug

x = 3

into

x^{4} - 18 x^{2} + 81 : 0

Minimum (3, 0)

Minimum (- 3, 0), Maximum (0, 81), Minimum (3, 0)

\frac{2}{( x - 4 ) ( - x - 6 )}

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

(- 10.2) (3.5)

x^{2} - x - ln (x)

f (x) = x^{\frac{7}{2}} - 3 x^{2}

What is the extreme f(x)=x^4-18x^2+81 ?
The extreme f(x)=x^4-18x^2+81 is Minimum(-3,0),Maximum(0,81),Minimum(3,0)