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

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

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

domain

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

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

Combine real regions and undefined points for final function domain

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

f (x) = \frac{1}{x + 9}

\frac{( x ^{2} + 8 x - 9 )}{x ^{2} + 3 x - 4}

2 x + 1

f (x) = x + 5

ln (x^{2} - 10 x)

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