What is the domain of f(x)=(sqrt(x-8))/(x-9) ?

The domain of f(x)=(sqrt(x-8))/(x-9) is 8 9

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domain of f(x)=(sqrt(x-8))/(x-9)

domain

f (x) = \frac{x - 8}{x - 9}

8 \leq x < 9 or x > 9

Interval Notation

[8, 9) \cup (9, \infty)

Solution steps

Find non-negative values for radicals:

x \geq 8

Find undefined (singularity) points:

x = 9

Combine real regions and undefined points for final function domain

8 \leq x < 9 or x > 9

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

h (x) = x + 15

\frac{7}{x ^{2} - 16} - \frac{( x + 4 )}{x ^{2} - 8 x + 16}

f (x) = \frac{5 x - 6}{5 x + 40}

x^{3} - 64

What is the domain of f(x)=(sqrt(x-8))/(x-9) ?
The domain of f(x)=(sqrt(x-8))/(x-9) is 8<= x<9\lor x>9