What is the critical (x^7-4x^6-16x^5+64x^4)/x ?

The critical (x^7-4x^6-16x^5+64x^4)/x is x=-(1+sqrt(73))/3 ,x=(-1+sqrt(73))/3 ,x=4

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critical (x^7-4x^6-16x^5+64x^4)/x

critical points

\frac{x ^{7} - 4 x ^{6} - 16 x ^{5} + 64 x ^{4}}{x}

x = - \frac{1 + 73}{3}, x = \frac{- 1 + 73}{3}, x = 4

Solution steps

\frac{x ^{7} - 4 x ^{6} - 16 x ^{5} + 64 x ^{4}}{x}

Find where

f^{'} (x)

is equal to zero or undefined

x = - \frac{1 + 73}{3}, x = 0, x = \frac{- 1 + 73}{3}, x = 4

Identify critical points not in

f (x)

domain

Domain of

\frac{x ^{7} - 4 x ^{6} - 16 x ^{5} + 64 x ^{4}}{x} : x < 0 or x > 0

\frac{x ^{7} - 4 x ^{6} - 16 x ^{5} + 64 x ^{4}}{x}

is not defined at

x = 0

, therefore:

x = - \frac{1 + 73}{3}, x = \frac{- 1 + 73}{3}, x = 4

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

x + 1 - x

f (x) = \frac{4 e ^{x}}{4 e ^{x} + 5}

\frac{4 x ^{2}}{x ^{2} - 4}

x^{3} + 3 x^{2} - 4

What is the critical (x^7-4x^6-16x^5+64x^4)/x ?
The critical (x^7-4x^6-16x^5+64x^4)/x is x=-(1+sqrt(73))/3 ,x=(-1+sqrt(73))/3 ,x=4