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

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

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

domain

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

Combine real regions and undefined points for final function domain

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

\frac{5 x - 4}{( x - 7 ) ^{2}}

f (x) = \frac{6 x ^{2} + 7}{8 x + 6}

y = \frac{x ^{2} + 6 x - 7}{x ^{2} + 25}

f (x) = x^{4} - 200 x^{2} + 10000

y = x + 6

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