What is the domain of y=sqrt((4x)/(6x^2+13x-5)) ?

The domain of y=sqrt((4x)/(6x^2+13x-5)) is -5/2 1/3

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domain of y=sqrt((4x)/(6x^2+13x-5))

domain

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

- \frac{5}{2} < x \leq 0 or x > \frac{1}{3}

Interval Notation

(- \frac{5}{2}, 0] \cup (\frac{1}{3}, \infty)

Solution steps

Find non-negative values for radicals:

- \frac{5}{2} < x \leq 0 or x > \frac{1}{3}

Find undefined (singularity) points:

x = \frac{1}{3}, x = - \frac{5}{2}

Combine real regions and undefined points for final function domain

- \frac{5}{2} < x \leq 0 or x > \frac{1}{3}

f (x) = x + 1 + x

g (x) = \frac{2 + x}{1 - x}

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

f (x) = 4 x - 6

f (x) = lo g_{3} (\frac{1}{x - 2})

What is the domain of y=sqrt((4x)/(6x^2+13x-5)) ?
The domain of y=sqrt((4x)/(6x^2+13x-5)) is -5/2 <x<= 0\lor x> 1/3