What is the domain of f(x)=sqrt((6+x)/(2+3x)) ?

The domain of f(x)=sqrt((6+x)/(2+3x)) is x -2/3

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domain of f(x)=sqrt((6+x)/(2+3x))

domain

f (x) = \frac{6 + x}{2 + 3 x}

x \leq - 6 or x > - \frac{2}{3}

Interval Notation

(- \infty, - 6] \cup (- \frac{2}{3}, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - 6 or x > - \frac{2}{3}

Find undefined (singularity) points:

x = - \frac{2}{3}

Combine real regions and undefined points for final function domain

x \leq - 6 or x > - \frac{2}{3}

f (x) = \frac{3 x + 5}{2 x ^{2} - 4 x - 6}

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

y = csc (x)

f (x) = \frac{x + 7}{x ^{2} - 49}

f (x) = x^{2} + 9 x + 2

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