What is the domain of (2x^2)/(sqrt(6x^2-x-2)) ?

The domain of (2x^2)/(sqrt(6x^2-x-2)) is x 2/3

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domain of (2x^2)/(sqrt(6x^2-x-2))

domain

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

x < - \frac{1}{2} or x > \frac{2}{3}

Interval Notation

(- \infty, - \frac{1}{2}) \cup (\frac{2}{3}, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - \frac{1}{2} or x \geq \frac{2}{3}

Find undefined (singularity) points:

x = \frac{2}{3}, x = - \frac{1}{2}

Combine real regions and undefined points for final function domain

x < - \frac{1}{2} or x > \frac{2}{3}

x^{9} - 6 x^{4} + 9

f (x) = 3 x + 24

y = \frac{7 x ^{2} - 63}{4}

x^{2} - y - 2 x + 1 = 0

f (x) = - \frac{7}{2} (5)^{x - 2} + 1

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