What is the domain of f(x)= x/(sqrt(3x^2+2x-8)) ?

The domain of f(x)= x/(sqrt(3x^2+2x-8)) is x 4/3

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domain of f(x)= x/(sqrt(3x^2+2x-8))

domain

f (x) = \frac{x}{3 x ^{2} + 2 x - 8}

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

Interval Notation

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

Solution steps

Find non-negative values for radicals:

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

Find undefined (singularity) points:

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

Combine real regions and undefined points for final function domain

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

\frac{x ^{3}}{x ^{4} - 1}

f (x) = 2 x + 8, - 3 < x < 2

\frac{x ^{2} - 36}{x + 7}

f (x) = - 4 x^{2} - 20 x - 21

f (x) = x^{2} + x, x \geq \frac{1}{2}

What is the domain of f(x)= x/(sqrt(3x^2+2x-8)) ?
The domain of f(x)= x/(sqrt(3x^2+2x-8)) is x<-2\lor x> 4/3