What is the domain of f(x)=(sqrt(2x+9))/(x^2-5x-6) ?

The domain of f(x)=(sqrt(2x+9))/(x^2-5x-6) is -9/2 6

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domain of f(x)=(sqrt(2x+9))/(x^2-5x-6)

domain

f (x) = \frac{2 x + 9}{x ^{2} - 5 x - 6}

- \frac{9}{2} \leq x < - 1 or - 1 < x < 6 or x > 6

Interval Notation

[- \frac{9}{2}, - 1) \cup (- 1, 6) \cup (6, \infty)

Solution steps

Find non-negative values for radicals:

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

Find undefined (singularity) points:

x = 6, x = - 1

Combine real regions and undefined points for final function domain

- \frac{9}{2} \leq x < - 1 or - 1 < x < 6 or x > 6

\frac{2}{x} - 1

x^{3} - 3 x^{2} + 3 x - 7

f (x) = x + \frac{34}{4}

\frac{- x + 10}{x ^{3}}

f (x) = \frac{7 - x}{x ^{2} - x - 2}

What is the domain of f(x)=(sqrt(2x+9))/(x^2-5x-6) ?
The domain of f(x)=(sqrt(2x+9))/(x^2-5x-6) is -9/2 <= x<-1\lor-1<x<6\lor x>6