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

The domain of f(x)= 1/(2x^2+x) is x 0

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

domain

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

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

Interval Notation

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

Solution steps

Find undefined (singularity) points:

x = 0, x = - \frac{1}{2}

The function domain

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

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

f (x) = lo g_{3} (2 x - 1) - 2

y = - 4

y = \frac{x ^{2} + x - 6}{x - 2}

y = (x + 3) (x - 1) (x - 4)

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