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

The domain of ((x/(2x^2-5)))/(sqrt(x)) is 0 sqrt(5/2)

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domain of ((x/(2x^2-5)))/(sqrt(x))

domain

\frac{( \frac{x}{2 x ^{2} - 5} )}{x}

0 < x < \frac{5}{2} or x > \frac{5}{2}

Interval Notation

(0, \frac{5}{2}) \cup (\frac{5}{2}, \infty)

Solution steps

Find non-negative values for radicals:

x \geq 0

Find undefined (singularity) points:

x = 0, x = \frac{5}{2}, x = - \frac{5}{2}

Combine real regions and undefined points for final function domain

0 < x < \frac{5}{2} or x > \frac{5}{2}

f (x) = - x^{3} + 3 x^{2} - 7

\frac{x + 3}{x - 5}

f (x) = y^{6} = x^{3} - 16 x

ln (e x)

What is the domain of ((x/(2x^2-5)))/(sqrt(x)) ?
The domain of ((x/(2x^2-5)))/(sqrt(x)) is 0<x<sqrt(5/2)\lor x>sqrt(5/2)