What is the domain of f(x)=(sqrt(x))/(2x-9) ?

The domain of f(x)=(sqrt(x))/(2x-9) is 0 9/2

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domain of f(x)=(sqrt(x))/(2x-9)

domain

f (x) = \frac{x}{2 x - 9}

0 \leq x < \frac{9}{2} or x > \frac{9}{2}

Interval Notation

[0, \frac{9}{2}) \cup (\frac{9}{2}, \infty)

Solution steps

Find non-negative values for radicals:

x \geq 0

Find undefined (singularity) points:

x = \frac{9}{2}

Combine real regions and undefined points for final function domain

0 \leq x < \frac{9}{2} or x > \frac{9}{2}

\frac{( x + 8 )}{( x ^{2} - 2 x - 15 )}

f (x) = \frac{1}{4 - 2 - x}

f (x) = - x^{2} - 3 x + 4

f (x) = \frac{x - 1}{x ^{2} + 6 x - 27}

y = \frac{x + 1}{2 x + 1}

What is the domain of f(x)=(sqrt(x))/(2x-9) ?
The domain of f(x)=(sqrt(x))/(2x-9) is 0<= x< 9/2 \lor x> 9/2