What is the critical f(x)=16x-4x^2 ?

The critical f(x)=16x-4x^2 is x=2

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critical f(x)=16x-4x^2

critical points

f (x) = 16 x - 4 x^{2}

x = 2

Solution steps

16 x - 4 x^{2}

Find where

f^{'} (x)

is equal to zero or undefined

x = 2

Identify critical points not in

f (x)

domain

Domain of

16 x - 4 x^{2} : - \infty < x < \infty

All critical points are in domain

x = 2

2 x - 3 y - 12 = 0

\frac{4 x - 1}{5 - x}

8 x - y = - 16, (0, 0)

\frac{2 x}{x - 6}

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