What is the domain of (sqrt(x))/(18x^2-33x-40) ?

The domain of (sqrt(x))/(18x^2-33x-40) is 0 8/3

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domain of (sqrt(x))/(18x^2-33x-40)

domain

\frac{x}{18 x ^{2} - 33 x - 40}

0 \leq x < \frac{8}{3} or x > \frac{8}{3}

Interval Notation

[0, \frac{8}{3}) \cup (\frac{8}{3}, \infty)

Solution steps

Find non-negative values for radicals:

x \geq 0

Find undefined (singularity) points:

x = \frac{8}{3}, x = - \frac{5}{6}

Combine real regions and undefined points for final function domain

0 \leq x < \frac{8}{3} or x > \frac{8}{3}

f (x) = x + 6 - 8

sin (arccos (x - 1))

f (x) = - x^{2} + 8 x - 8

f (x) = \frac{( x 5 ^{x} )}{5 ^{(x + 3)} - 6}

f (x) = \frac{( a , b )}{a} b = 12

What is the domain of (sqrt(x))/(18x^2-33x-40) ?
The domain of (sqrt(x))/(18x^2-33x-40) is 0<= x< 8/3 \lor x> 8/3