What is the domain of (sqrt(x))/(5x^2+4x-1) ?

The domain of (sqrt(x))/(5x^2+4x-1) is 0 1/5

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domain of (sqrt(x))/(5x^2+4x-1)

domain

\frac{x}{5 x ^{2} + 4 x - 1}

0 \leq x < \frac{1}{5} or x > \frac{1}{5}

Interval Notation

[0, \frac{1}{5}) \cup (\frac{1}{5}, \infty)

Solution steps

Find non-negative values for radicals:

x \geq 0

Find undefined (singularity) points:

x = \frac{1}{5}, x = - 1

Combine real regions and undefined points for final function domain

0 \leq x < \frac{1}{5} or x > \frac{1}{5}

x^{4} - 4 x^{2}

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

g (t) = - \frac{9}{2 t ^{\frac{3}{2}}}

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

x^{4} - 12 x^{3}

What is the domain of (sqrt(x))/(5x^2+4x-1) ?
The domain of (sqrt(x))/(5x^2+4x-1) is 0<= x< 1/5 \lor x> 1/5