What is the critical f(x)= 1/(x^3-6x^2+1) ?

The critical f(x)= 1/(x^3-6x^2+1) is x=0,x=4

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critical f(x)= 1/(x^3-6x^2+1)

critical points

f (x) = \frac{1}{x ^{3} - 6 x ^{2} + 1}

x = 0, x = 4

Solution steps

\frac{1}{x ^{3} - 6 x ^{2} + 1}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = 4, x \approx - 0.39542 \dots, x \approx 0.42346 \dots, x = 5.97196 \dots

Identify critical points not in

f (x)

domain

Domain of

\frac{1}{x ^{3} - 6 x ^{2} + 1} : x < - 0.39542 \dots or - 0.39542 \dots < x < 0.42346 \dots or 0.42346 \dots < x < 5.97196 \dots or x > 5.97196 \dots

\frac{1}{x ^{3} - 6 x ^{2} + 1}

is not defined at

x \approx - 0.39542 \dots, x \approx 0.42346 \dots, x = 5.97196 \dots

, therefore:

x = 0, x = 4

f (x) = x^{2} + 6 x

\frac{x ^{2} - 24}{x - 5}

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

y = \frac{x ^{2} - 3 x + 2}{x ^{2} - 4 x + 3}

f (x) = - 7 x^{3} + 21 x + 2

What is the critical f(x)= 1/(x^3-6x^2+1) ?
The critical f(x)= 1/(x^3-6x^2+1) is x=0,x=4