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

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

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

domain

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

0 \leq x < \frac{1}{2} or x > \frac{1}{2}

Interval Notation

[0, \frac{1}{2}) \cup (\frac{1}{2}, \infty)

Solution steps

Find non-negative values for radicals:

x \geq 0

Find undefined (singularity) points:

x = \frac{1}{2}, x = - 1

Combine real regions and undefined points for final function domain

0 \leq x < \frac{1}{2} or x > \frac{1}{2}

m = - \frac{3}{4}, (- 5, 3)

x \cdot e^{\frac{4}{x}} + 3

\frac{1}{x} + 1

f (x) = - x^{2} - 2 x - 5

f (x) = lo g_{4} (x)

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