What is the domain of f(x)=(sqrt(x+1))/(x-8) ?

The domain of f(x)=(sqrt(x+1))/(x-8) is -1 8

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domain of f(x)=(sqrt(x+1))/(x-8)

domain

f (x) = \frac{x + 1}{x - 8}

- 1 \leq x < 8 or x > 8

Interval Notation

[- 1, 8) \cup (8, \infty)

Solution steps

Find non-negative values for radicals:

x \geq - 1

Find undefined (singularity) points:

x = 8

Combine real regions and undefined points for final function domain

- 1 \leq x < 8 or x > 8

f (x) = 6 x^{2} - x - 12

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

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

\frac{5 x - 10}{27 - 6 x}

y = \frac{2}{5} x + 1, (10, - 8)

What is the domain of f(x)=(sqrt(x+1))/(x-8) ?
The domain of f(x)=(sqrt(x+1))/(x-8) is -1<= x<8\lor x>8