What is the domain of (sqrt(6-x))/(x+9) ?

The domain of (sqrt(6-x))/(x+9) is x<-9\lor-9<x<= 6

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domain of (sqrt(6-x))/(x+9)

domain

\frac{6 - x}{x + 9}

x < - 9 or - 9 < x \leq 6

Interval Notation

(- \infty, - 9) \cup (- 9, 6]

Solution steps

Find non-negative values for radicals:

x \leq 6

Find undefined (singularity) points:

x = - 9

Combine real regions and undefined points for final function domain

x < - 9 or - 9 < x \leq 6

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

f (x) = \frac{1}{25 - x ^{2}}

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

f (x) = lo g_{10} (x^{2} - 9)

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

What is the domain of (sqrt(6-x))/(x+9) ?
The domain of (sqrt(6-x))/(x+9) is x<-9\lor-9<x<= 6