What is the domain of f(x)=sqrt((2x^2-x-1)/(x^2+3x)) ?

The domain of f(x)=sqrt((2x^2-x-1)/(x^2+3x)) is x = 1

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domain of f(x)=sqrt((2x^2-x-1)/(x^2+3x))

domain

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

x < - 3 or - \frac{1}{2} \leq x < 0 or x \geq 1

Interval Notation

(- \infty, - 3) \cup [- \frac{1}{2}, 0) \cup [1, \infty)

Solution steps

Find non-negative values for radicals:

x < - 3 or - \frac{1}{2} \leq x < 0 or x \geq 1

Find undefined (singularity) points:

x = 0, x = - 3

Combine real regions and undefined points for final function domain

x < - 3 or - \frac{1}{2} \leq x < 0 or x \geq 1

f (x) = \frac{x + 6}{2 x ^{2} + 24 x + 70}

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

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

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

y = arcsin (2 x)

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