What is the domain of f(x)=(sqrt(x))/(6x^2+5x-1) ?

The domain of f(x)=(sqrt(x))/(6x^2+5x-1) is 0 1/6

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domain of f(x)=(sqrt(x))/(6x^2+5x-1)

domain

f (x) = \frac{x}{6 x ^{2} + 5 x - 1}

0 \leq x < \frac{1}{6} or x > \frac{1}{6}

Interval Notation

[0, \frac{1}{6}) \cup (\frac{1}{6}, \infty)

Solution steps

Find non-negative values for radicals:

x \geq 0

Find undefined (singularity) points:

x = \frac{1}{6}, x = - 1

Combine real regions and undefined points for final function domain

0 \leq x < \frac{1}{6} or x > \frac{1}{6}

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

y = 5 x - 9

f (x) = x^{2} + 5 x + 6

\frac{x ^{2} - 4}{x ^{3}}

\frac{3 - 2 x}{3 x + 4}

What is the domain of f(x)=(sqrt(x))/(6x^2+5x-1) ?
The domain of f(x)=(sqrt(x))/(6x^2+5x-1) is 0<= x< 1/6 \lor x> 1/6