What is the critical g(x)=(x^3)/((x+1)) ?

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

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critical g(x)=(x^3)/((x+1))

critical points

g (x) = \frac{x ^{3}}{( x + 1 )}

x = - \frac{3}{2}, x = 0

Solution steps

\frac{x ^{3}}{x + 1}

Find where

f^{'} (x)

is equal to zero or undefined

x = - \frac{3}{2}, x = 0, x = - 1

Identify critical points not in

f (x)

domain

Domain of

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

\frac{x ^{3}}{x + 1}

is not defined at

x = - 1

, therefore:

x = - \frac{3}{2}, x = 0

\frac{4}{x + 3}

y - 5 = - \frac{1}{5} (x - 1)

x^{2} + x + 2

- \frac{1}{4} cos (- \frac{1}{4} x)

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

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