What is the domain of f(x)=(sqrt(2x-5))/(x^2-6x) ?

The domain of f(x)=(sqrt(2x-5))/(x^2-6x) is 5/2 6

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domain of f(x)=(sqrt(2x-5))/(x^2-6x)

domain

f (x) = \frac{2 x - 5}{x ^{2} - 6 x}

\frac{5}{2} \leq x < 6 or x > 6

Interval Notation

[\frac{5}{2}, 6) \cup (6, \infty)

Solution steps

Find non-negative values for radicals:

x \geq \frac{5}{2}

Find undefined (singularity) points:

x = 6, x = 0

Combine real regions and undefined points for final function domain

\frac{5}{2} \leq x < 6 or x > 6

y = \frac{x}{1 - ln ( x - 8 )}

f (x) = - (x - 1)^{3} + 27

3 e^{- x} - 2

3 - 2 lo g_{10} (x + 3)

f (x) = ∣ 2 x - 19 ∣

What is the domain of f(x)=(sqrt(2x-5))/(x^2-6x) ?
The domain of f(x)=(sqrt(2x-5))/(x^2-6x) is 5/2 <= x<6\lor x>6