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

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

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

domain

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

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

Interval Notation

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

Solution steps

Find non-negative values for radicals:

x \geq 2

Find undefined (singularity) points:

x = \frac{5}{2}

Combine real regions and undefined points for final function domain

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

\frac{3 x - 5}{2 x + 1}

8 x^{4} + 24 x^{2} + 21

f (x) = \frac{x + 1}{x + 2}

(\frac{y - 1}{2 y})^{2}

f (x) = (x + 2)^{\frac{1}{2}} + 3

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