What is the critical f(x)=x^4-8x^2+9 ?

The critical f(x)=x^4-8x^2+9 is x=-2,x=0,x=2

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critical f(x)=x^4-8x^2+9

critical points

f (x) = x^{4} - 8 x^{2} + 9

x = - 2, x = 0, x = 2

Solution steps

x^{4} - 8 x^{2} + 9

Find where

f^{'} (x)

is equal to zero or undefined

x = - 2, x = 0, x = 2

Identify critical points not in

f (x)

domain

Domain of

x^{4} - 8 x^{2} + 9 : - \infty < x < \infty

All critical points are in domain

x = - 2, x = 0, x = 2

(x + 3) e^{- 2 x}

10000 - x^{3} + 33 x^{2} + 600 x

\frac{x ( x + 4 )}{( x - 1 ) 2}

y = - 2 x + 3

f (x) = \frac{6 x ^{2} + x - 9}{x ^{2} + 4}

What is the critical f(x)=x^4-8x^2+9 ?
The critical f(x)=x^4-8x^2+9 is x=-2,x=0,x=2