What is the domain of f(x)= 1/(x^2+2x-3)+sqrt(x+4) ?

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

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domain of f(x)= 1/(x^2+2x-3)+sqrt(x+4)

domain

f (x) = \frac{1}{x ^{2} + 2 x - 3} + x + 4

- 4 \leq x < - 3 or - 3 < x < 1 or x > 1

Interval Notation

[- 4, - 3) \cup (- 3, 1) \cup (1, \infty)

Solution steps

Find non-negative values for radicals:

x \geq - 4

Find undefined (singularity) points:

x = 1, x = - 3

Combine real regions and undefined points for final function domain

- 4 \leq x < - 3 or - 3 < x < 1 or x > 1

\frac{6 + x}{1 - 6 x}

r (x) = \frac{x ^{2} + 8 x + 18}{2 x ^{2} + 16 x + 32}

f (x) = 36 - x^{2} - 3

\frac{1}{x 1 - x ^{2}}

\frac{5 - 2 x}{4 x - 1}

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