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

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

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

domain

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

- 2 \leq x < - 1 or - 1 < x < 2 or x > 2

Interval Notation

[- 2, - 1) \cup (- 1, 2) \cup (2, \infty)

Solution steps

Find non-negative values for radicals:

x \geq - 2

Find undefined (singularity) points:

x = 2, x = - 1

Combine real regions and undefined points for final function domain

- 2 \leq x < - 1 or - 1 < x < 2 or x > 2

f (x) = x^{2} - 4 x + 9

f (x) = \frac{1}{x ^{2} - 8 x + 16}

y = x^{2} - x

f (x) = - x^{2} + 2 x

f (x) = arccos (x - 3)

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