What is the domain of f(x)= 1/(x^2(x+9)) ?

The domain of f(x)= 1/(x^2(x+9)) is x 0

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domain of f(x)= 1/(x^2(x+9))

domain

f (x) = \frac{1}{x ^{2} ( x + 9 )}

x < - 9 or - 9 < x < 0 or x > 0

Interval Notation

(- \infty, - 9) \cup (- 9, 0) \cup (0, \infty)

Solution steps

Find undefined (singularity) points:

x = 0, x = - 9

The function domain

x < - 9 or - 9 < x < 0 or x > 0

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

f (x) = 2 x + 3 y - 5 = 0

f (x) = \frac{3 x}{x - 3}

f (x) = x^{2} - 3

f (x) = 8 + x

What is the domain of f(x)= 1/(x^2(x+9)) ?
The domain of f(x)= 1/(x^2(x+9)) is x<-9\lor-9<x<0\lor x>0