What is the domain of f(x)=(sqrt(x+39))/(x-3) ?

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

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domain of f(x)=(sqrt(x+39))/(x-3)

domain

f (x) = \frac{x + 39}{x - 3}

- 39 \leq x < 3 or x > 3

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

f (x) = \frac{x + 5}{x ^{2} - 25}

y = \frac{1}{3} x + 4

f (x) = x - 1 + 1

2 x + 3 y = 5

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

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