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

The domain of y=(sqrt(x+1))/(x^2) is -1 0

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domain of y=(sqrt(x+1))/(x^2)

domain

y = \frac{x + 1}{x ^{2}}

- 1 \leq x < 0 or x > 0

Interval Notation

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

Solution steps

Find non-negative values for radicals:

x \geq - 1

Find undefined (singularity) points:

x = 0

Combine real regions and undefined points for final function domain

- 1 \leq x < 0 or x > 0

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

f (x) = 8 x - 16

f (x) = (- 4.2)

h (x) = - \frac{1}{x - 6} + 5

\frac{x ^{3} - 64}{x ^{2} - 5 x + 4}

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