What is the domain of f(x)= 1/(sqrt(6x^2-x-12)) ?

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

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domain of f(x)= 1/(sqrt(6x^2-x-12))

domain

f (x) = \frac{1}{6 x ^{2} - x - 12}

x < - \frac{4}{3} or x > \frac{3}{2}

Interval Notation

(- \infty, - \frac{4}{3}) \cup (\frac{3}{2}, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - \frac{4}{3} or x \geq \frac{3}{2}

Find undefined (singularity) points:

x = \frac{3}{2}, x = - \frac{4}{3}

Combine real regions and undefined points for final function domain

x < - \frac{4}{3} or x > \frac{3}{2}

\frac{4 x + 7}{6 x ^{2} + 13 x - 5}

\frac{3 x - 5}{2 - 4 x}

f (x) = \frac{5 x + 2}{7}

f (x) = 2 x^{4}, x \geq 0

f (x) = - 4 + lo g_{2} (5 - 2 x)

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