What is the domain of f(x)=sqrt(1+9/x) ?

The domain of f(x)=sqrt(1+9/x) is x 0

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domain of f(x)=sqrt(1+9/x)

domain

f (x) = 1 + \frac{9}{x}

x \leq - 9 or x > 0

Interval Notation

(- \infty, - 9] \cup (0, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - 9 or x > 0

Find undefined (singularity) points:

x = 0

Combine real regions and undefined points for final function domain

x \leq - 9 or x > 0

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

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

5 + (10 + x)^{\frac{1}{2}}

f (x) = \frac{( 5 x ^{2} - 3 )}{x + 2}

f (x) = \frac{5 + 2 ^{x}}{1 - 2 ^{x}}

What is the domain of f(x)=sqrt(1+9/x) ?
The domain of f(x)=sqrt(1+9/x) is x<=-9\lor x>0