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

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

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

domain

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

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

Interval Notation

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

Solution steps

Find non-negative values for radicals:

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

Find undefined (singularity) points:

x = 1, x = - 3

Combine real regions and undefined points for final function domain

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

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

tan^{\frac{1}{2}} (x) d x

f (x) = x^{3} - 8

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

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

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