What is the domain of (sqrt(9-x^2))/(3x-1) ?

The domain of (sqrt(9-x^2))/(3x-1) is -3<= x< 1/3 \lor 1/3 <x<= 3

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domain of (sqrt(9-x^2))/(3x-1)

domain

\frac{9 - x ^{2}}{3 x - 1}

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

Interval Notation

Solution steps

Find non-negative values for radicals:

- 3 \leq x \leq 3

Find undefined (singularity) points:

x = \frac{1}{3}

Combine real regions and undefined points for final function domain

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

ln (x^{2} - 4) + π

y = \frac{x ^{2}}{x - 6}

(x^{2} - 4 x - 5) - (x^{2} + 3 x + 2)

f (x) = 12 (x - 8) (x - 6)

f (x) = \frac{x + 3}{\frac{x + 4}{2}}

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