What is the domain of f(x)=(1-6x+9x^2)/(3x-1) ?

The domain of f(x)=(1-6x+9x^2)/(3x-1) is x 1/3

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domain of f(x)=(1-6x+9x^2)/(3x-1)

domain

f (x) = \frac{1 - 6 x + 9 x ^{2}}{3 x - 1}

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

Interval Notation

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

Solution steps

Find undefined (singularity) points:

x = \frac{1}{3}

The function domain

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

f (x) = \frac{4 x + 10}{x + 3}

f (x) = - x^{2} - 16

f (x) = 5 x^{2} + 11 g (x) = x + 2

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

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

What is the domain of f(x)=(1-6x+9x^2)/(3x-1) ?
The domain of f(x)=(1-6x+9x^2)/(3x-1) is x< 1/3 \lor x> 1/3