What is the domain of f(x)= 1/(sqrt(\frac{x){x+1)}} ?

The domain of f(x)= 1/(sqrt(\frac{x){x+1)}} is x 0

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domain of f(x)= 1/(sqrt(\frac{x){x+1)}}

domain

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

x < - 1 or x > 0

Interval Notation

(- \infty, - 1) \cup (0, \infty)

Solution steps

Find non-negative values for radicals:

x < - 1 or x \geq 0

Find undefined (singularity) points:

x = 0, x = - 1

Combine real regions and undefined points for final function domain

x < - 1 or x > 0

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

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

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}}

What is the domain of f(x)= 1/(sqrt(\frac{x){x+1)}} ?
The domain of f(x)= 1/(sqrt(\frac{x){x+1)}} is x<-1\lor x>0