What is the domain of f(x)=(sqrt(2x+9))/(x-2) ?

The domain of f(x)=(sqrt(2x+9))/(x-2) is -9/2 2

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domain of f(x)=(sqrt(2x+9))/(x-2)

domain

f (x) = \frac{2 x + 9}{x - 2}

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

Interval Notation

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

Solution steps

Find non-negative values for radicals:

x \geq - \frac{9}{2}

Find undefined (singularity) points:

x = 2

Combine real regions and undefined points for final function domain

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

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

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

f (x) = \frac{16 - x ^{2}}{x + 3}

2 x - 3 y = 8

f (x) = 6 tan (0.2 x)

What is the domain of f(x)=(sqrt(2x+9))/(x-2) ?
The domain of f(x)=(sqrt(2x+9))/(x-2) is -9/2 <= x<2\lor x>2