What is the critical x^4-6x^3 ?

The critical x^4-6x^3 is x=0,x= 9/2

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critical x^4-6x^3

critical points

x^{4} - 6 x^{3}

x = 0, x = \frac{9}{2}

Solution steps

x^{4} - 6 x^{3}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = \frac{9}{2}

Identify critical points not in

f (x)

domain

Domain of

x^{4} - 6 x^{3} : - \infty < x < \infty

All critical points are in domain

x = 0, x = \frac{9}{2}

(x + 5)^{2} + 3

y = 1

f (x) = \frac{8}{x - 6}

y = 4 - x^{2}

f (x) = \frac{( x + 4 )}{( 2 x - 7 )}