What is the domain of f(x)= 1/((x-3)(3x+1)) ?

The domain of f(x)= 1/((x-3)(3x+1)) is x 3

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domain of f(x)= 1/((x-3)(3x+1))

domain

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

x < - \frac{1}{3} or - \frac{1}{3} < x < 3 or x > 3

Interval Notation

(- \infty, - \frac{1}{3}) \cup (- \frac{1}{3}, 3) \cup (3, \infty)

Solution steps

Find undefined (singularity) points:

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

The function domain

x < - \frac{1}{3} or - \frac{1}{3} < x < 3 or x > 3

f (x) = \frac{3 x + 4}{( x + 4 ) ( x + 3 )}

f (x, y) = \frac{1}{x}

\frac{1}{2 x + 2}

f (x) = 3 - (x + 1)^{2}, - 3 < x \leq 1

x (\frac{4 x + 1976}{x - 6})

What is the domain of f(x)= 1/((x-3)(3x+1)) ?
The domain of f(x)= 1/((x-3)(3x+1)) is x<-1/3 \lor-1/3 <x<3\lor x>3