What is the domain of f(x)=(sqrt(2x-5))/(3x-9) ?

The domain of f(x)=(sqrt(2x-5))/(3x-9) is 5/2 3

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domain of f(x)=(sqrt(2x-5))/(3x-9)

domain

f (x) = \frac{2 x - 5}{3 x - 9}

\frac{5}{2} \leq x < 3 or x > 3

Interval Notation

[\frac{5}{2}, 3) \cup (3, \infty)

Solution steps

Find non-negative values for radicals:

x \geq \frac{5}{2}

Find undefined (singularity) points:

x = 3

Combine real regions and undefined points for final function domain

\frac{5}{2} \leq x < 3 or x > 3

y = - \frac{3}{5} x + 7

y = x^{3} - x

f (x) = \frac{( x ^{2} - 25 )}{x ^{2} - 4 x - 5}

f (x) = cos (x)

1200 + 500 sin (0.3 t)

What is the domain of f(x)=(sqrt(2x-5))/(3x-9) ?
The domain of f(x)=(sqrt(2x-5))/(3x-9) is 5/2 <= x<3\lor x>3