What is the domain of ((sqrt(x)+3))/(x^2+4x-5) ?

The domain of ((sqrt(x)+3))/(x^2+4x-5) is 0 1

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domain of ((sqrt(x)+3))/(x^2+4x-5)

domain

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

0 \leq x < 1 or x > 1

Interval Notation

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

Solution steps

Find non-negative values for radicals:

x \geq 0

Find undefined (singularity) points:

x = 1, x = - 5

Combine real regions and undefined points for final function domain

0 \leq x < 1 or x > 1

f (x) = \frac{ln ( x ^{2} - 9 )}{x ^{2} - 16}

f (x) = 2 x^{3} + 4 x^{2} + 2

(x - 14)^{2}, x \leq 14

f (x) = \frac{5 x}{7 x - 9}

f (x) = \frac{x - 7}{x ^{2} - 2 x - 35}

What is the domain of ((sqrt(x)+3))/(x^2+4x-5) ?
The domain of ((sqrt(x)+3))/(x^2+4x-5) is 0<= x<1\lor x>1