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

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

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

domain

f (x) = \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}

f (x) = 5 - 2 x + 4

f (x) = \frac{x + 5}{x ^{2} - 9}

f (x) = 2 x + 3, - 2 \leq x < 1

f (x) = x - 4 x^{3}

f (x) = \frac{( 4 x + 3 )}{( 2 x - 1 )}

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