What is the domain of f(x)=(2x)/(sqrt(3x^2-5)) ?

The domain of f(x)=(2x)/(sqrt(3x^2-5)) is x sqrt(5/3)

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domain of f(x)=(2x)/(sqrt(3x^2-5))

domain

f (x) = \frac{2 x}{3 x ^{2} - 5}

x < - \frac{5}{3} or x > \frac{5}{3}

Interval Notation

(- \infty, - \frac{5}{3}) \cup (\frac{5}{3}, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - \frac{5}{3} or x \geq \frac{5}{3}

Find undefined (singularity) points:

x = \frac{5}{3}, x = - \frac{5}{3}

Combine real regions and undefined points for final function domain

x < - \frac{5}{3} or x > \frac{5}{3}

ln (4 - x)

f (x) = - e^{8 - x}

f (x) = \frac{x + 4}{24 - x ^{2} - 49}

0.2 x^{2}

f (x) = \frac{5 + x}{4 - 2 x}

What is the domain of f(x)=(2x)/(sqrt(3x^2-5)) ?
The domain of f(x)=(2x)/(sqrt(3x^2-5)) is x<-sqrt(5/3)\lor x>sqrt(5/3)