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

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

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

critical points

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

x = - 3, x = 0, x = 3

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = - 3, x = 0, x = 3

Identify critical points not in

f (x)

domain

Domain of

x^{4} - 18 x^{2} + 81 : - \infty < x < \infty

All critical points are in domain

x = - 3, x = 0, x = 3

f (x) = 2 cos (3 x + \frac{π}{2})

(- 3.3) (5.1)

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

- \frac{x ^{2}}{2} - 3 x - \frac{1}{2}

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

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