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

The domain of f(x)=sqrt((x^2-3x+2)/(2x^2-x)) is x = 2

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domain of f(x)=sqrt((x^2-3x+2)/(2x^2-x))

domain

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

x < 0 or \frac{1}{2} < x \leq 1 or x \geq 2

Interval Notation

(- \infty, 0) \cup (\frac{1}{2}, 1] \cup [2, \infty)

Solution steps

Find non-negative values for radicals:

x < 0 or \frac{1}{2} < x \leq 1 or x \geq 2

Find undefined (singularity) points:

x = \frac{1}{2}, x = 0

Combine real regions and undefined points for final function domain

x < 0 or \frac{1}{2} < x \leq 1 or x \geq 2

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

5 x^{2}

f (x) = \frac{2}{x ^{2} + 2 x - 35}

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

What is the domain of f(x)=sqrt((x^2-3x+2)/(2x^2-x)) ?
The domain of f(x)=sqrt((x^2-3x+2)/(2x^2-x)) is x<0\lor 1/2 <x<= 1\lor x>= 2