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

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

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

domain

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

x < - \frac{1}{2} or - \frac{1}{2} < x < \frac{1}{2} or x > \frac{1}{2}

Interval Notation

(- \infty, - \frac{1}{2}) \cup (- \frac{1}{2}, \frac{1}{2}) \cup (\frac{1}{2}, \infty)

Solution steps

Find undefined (singularity) points:

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

The function domain

x < - \frac{1}{2} or - \frac{1}{2} < x < \frac{1}{2} or x > \frac{1}{2}

\frac{x ^{3} + 3 x ^{2} + 2 x}{3 x ^{3} + 3 x ^{2} - 6 x}

f (x) = - \frac{1}{2} x^{2} - 4 x + 10

f (x) = (x - 13)^{2}, x \leq 13

y = 3 sin (2 x)

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