What is the domain of f(x)= 1/(2x^2+5x-3) ?

The domain of f(x)= 1/(2x^2+5x-3) is x 1/2

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domain of f(x)= 1/(2x^2+5x-3)

domain

f (x) = \frac{1}{2 x ^{2} + 5 x - 3}

x < - 3 or - 3 < x < \frac{1}{2} or x > \frac{1}{2}

Interval Notation

(- \infty, - 3) \cup (- 3, \frac{1}{2}) \cup (\frac{1}{2}, \infty)

Solution steps

Find undefined (singularity) points:

x = \frac{1}{2}, x = - 3

The function domain

x < - 3 or - 3 < x < \frac{1}{2} or x > \frac{1}{2}

y = 4 x - 1

f (x) = x^{2} - 5 x + 7

- \frac{1 - 3 x}{x - 1}

2 (x + 3)^{2} - 4

f (x) = 7 x^{4} - 5 x^{3} - 4

What is the domain of f(x)= 1/(2x^2+5x-3) ?
The domain of f(x)= 1/(2x^2+5x-3) is x<-3\lor-3<x< 1/2 \lor x> 1/2