What is the domain of f(x)= 1/(sqrt(2x^2-5x)) ?

The domain of f(x)= 1/(sqrt(2x^2-5x)) is x 5/2

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domain of f(x)= 1/(sqrt(2x^2-5x))

domain

f (x) = \frac{1}{2 x ^{2} - 5 x}

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

Interval Notation

(- \infty, 0) \cup (\frac{5}{2}, \infty)

Solution steps

Find non-negative values for radicals:

x \leq 0 or x \geq \frac{5}{2}

Find undefined (singularity) points:

x = \frac{5}{2}, x = 0

Combine real regions and undefined points for final function domain

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

f (x) = 6 x^{2} + 20 x^{3} + 13 x^{2} - 4 x

g (x) = \frac{x - 1}{x ^{2} - 8 x - 9}

f (x) = 2^{- x} - 2

f (x) = 2^{- x} - 1

What is the domain of f(x)= 1/(sqrt(2x^2-5x)) ?
The domain of f(x)= 1/(sqrt(2x^2-5x)) is x<0\lor x> 5/2