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

The critical f(x)=x^2-8x is x=4

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critical f(x)=x^2-8x

critical points

f (x) = x^{2} - 8 x

x = 4

Solution steps

x^{2} - 8 x

Find where

f^{'} (x)

is equal to zero or undefined

x = 4

Identify critical points not in

f (x)

domain

Domain of

x^{2} - 8 x : - \infty < x < \infty

All critical points are in domain

x = 4

f (x) = - \frac{1}{x + 3}

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

f (x) = 3 cos (x)

f (x) = 8 - 5 x^{2}

f (x) = - x^{3} + 12 x - 16