What is the critical ((x^3))/(x^2-1) ?

The critical ((x^3))/(x^2-1) is x=-sqrt(3),x=0,x=sqrt(3)

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critical ((x^3))/(x^2-1)

critical points

\frac{( x ^{3} )}{x ^{2} - 1}

x = - 3, x = 0, x = 3

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = - 3, x = 0, x = 3, x = - 1, x = 1

Identify critical points not in

f (x)

domain

Domain of

\frac{x ^{3}}{x ^{2} - 1} : x < - 1 or - 1 < x < 1 or x > 1

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

is not defined at

x = - 1, x = 1

, therefore:

x = - 3, x = 0, x = 3

(7 x + 5) e^{- 6 x}

f (x) = x^{4} - x^{2} - 2, - 2 \leq x \leq 2

f (x) = x^{6} e^{- 8 x}

f (x) = 8 x^{2} - x^{4}

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

What is the critical ((x^3))/(x^2-1) ?
The critical ((x^3))/(x^2-1) is x=-sqrt(3),x=0,x=sqrt(3)