What is the domain of f(x)=((sqrt(x)))/(4x^2+3x-1) ?

The domain of f(x)=((sqrt(x)))/(4x^2+3x-1) is 0 1/4

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domain of f(x)=((sqrt(x)))/(4x^2+3x-1)

domain

f (x) = \frac{( x )}{4 x ^{2} + 3 x - 1}

0 \leq x < \frac{1}{4} or x > \frac{1}{4}

Interval Notation

[0, \frac{1}{4}) \cup (\frac{1}{4}, \infty)

Solution steps

Find non-negative values for radicals:

x \geq 0

Find undefined (singularity) points:

x = \frac{1}{4}, x = - 1

Combine real regions and undefined points for final function domain

0 \leq x < \frac{1}{4} or x > \frac{1}{4}

3

f (x) = \frac{3 x ^{4} + 19}{15 x ^{5} + 25 x}

f (x) = \frac{7 x + 9}{9 x - 7} \cdot \frac{9 x}{9 x - 7}

1 - x^{2}

What is the domain of f(x)=((sqrt(x)))/(4x^2+3x-1) ?
The domain of f(x)=((sqrt(x)))/(4x^2+3x-1) is 0<= x< 1/4 \lor x> 1/4