What is the domain of (sqrt(9-2x))/(x^2-49) ?

The domain of (sqrt(9-2x))/(x^2-49) is x<-7\lor-7<x<= 9/2

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domain of (sqrt(9-2x))/(x^2-49)

domain

\frac{9 - 2 x}{x ^{2} - 49}

x < - 7 or - 7 < x \leq \frac{9}{2}

Interval Notation

(- \infty, - 7) \cup (- 7, \frac{9}{2}]

Solution steps

Find non-negative values for radicals:

x \leq \frac{9}{2}

Find undefined (singularity) points:

x = 7, x = - 7

Combine real regions and undefined points for final function domain

x < - 7 or - 7 < x \leq \frac{9}{2}

\frac{x + 4}{x + 3}

f (x) = ln (x - 3) - 1

\frac{3 x}{4 - x}

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

f (x) = \frac{1}{9.5 x - 6.1 x ^{2}}

What is the domain of (sqrt(9-2x))/(x^2-49) ?
The domain of (sqrt(9-2x))/(x^2-49) is x<-7\lor-7<x<= 9/2