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

The domain of (2x+3)/(2x^2+x-1) is x 1/2

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domain of (2x+3)/(2x^2+x-1)

domain

\frac{2 x + 3}{2 x ^{2} + x - 1}

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

Interval Notation

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

Solution steps

Find undefined (singularity) points:

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

The function domain

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

f (x) = (\frac{7 x}{x - 5})

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

f (x) = \frac{x ^{2} + 2 x - 3}{x ^{2} + 5 x + 6}

f (x) = \frac{3}{x - 3}

7 x^{2} + 8

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