What is the domain of f(x)=(x+1)/(2x+sqrt(39+x)) ?

The domain of f(x)=(x+1)/(2x+sqrt(39+x)) is -39 -3

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domain of f(x)=(x+1)/(2x+sqrt(39+x))

domain

f (x) = \frac{x + 1}{2 x + 39 + x}

- 39 \leq x < - 3 or x > - 3

Interval Notation

[- 39, - 3) \cup (- 3, \infty)

Solution steps

Find non-negative values for radicals:

x \geq - 39

Find undefined (singularity) points:

x = - 3

Combine real regions and undefined points for final function domain

- 39 \leq x < - 3 or x > - 3

x^{3} (x + 5)^{2} + 5

ln (tan (x) + sec (x))

f (x) = \frac{1}{x - 6}

\frac{6 + x ^{4}}{x ^{2} - x ^{4}}

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

What is the domain of f(x)=(x+1)/(2x+sqrt(39+x)) ?
The domain of f(x)=(x+1)/(2x+sqrt(39+x)) is -39<= x<-3\lor x>-3