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

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

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

domain

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

x \geq \frac{9}{2}

Interval Notation

[\frac{9}{2}, \infty)

Solution steps

Find non-negative values for radicals:

x \geq \frac{9}{2}

Find undefined (singularity) points:

x = - \frac{9}{2}

Combine real regions and undefined points for final function domain

x \geq \frac{9}{2}

lo g_{4} (x + 1) - 2

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

lo g_{4} (x + 1) + 2

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

\frac{t - 2}{t + 1}

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