What is the domain of f(x)=sqrt((3-12x)/(6+4x)) ?

The domain of f(x)=sqrt((3-12x)/(6+4x)) is -3/2 <x<= 1/4

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domain of f(x)=sqrt((3-12x)/(6+4x))

domain

f (x) = \frac{3 - 12 x}{6 + 4 x}

- \frac{3}{2} < x \leq \frac{1}{4}

Interval Notation

(- \frac{3}{2}, \frac{1}{4}]

Solution steps

Find non-negative values for radicals:

- \frac{3}{2} < x \leq \frac{1}{4}

Find undefined (singularity) points:

x = - \frac{3}{2}

Combine real regions and undefined points for final function domain

- \frac{3}{2} < x \leq \frac{1}{4}

\frac{x + 4}{x - 9}

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

\frac{( x ^{3} - 8 )}{( x ^{2} - 5 x + 6 )}

3 (3 x + 5) + 5

\frac{1}{x - 4} + \frac{1}{6 - x}

What is the domain of f(x)=sqrt((3-12x)/(6+4x)) ?
The domain of f(x)=sqrt((3-12x)/(6+4x)) is -3/2 <x<= 1/4