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

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

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

domain

g (x) = \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}

f (x) = \frac{x ^{2} - 4}{4 x - 16}

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

f (x) = 3 - 4 x

- 23.6 (1000 sin (x) - 225) + 822.2

y = (x - 7)^{3}

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