What is the domain of f(x)= 2/(6x^2+13x-5) ?

The domain of f(x)= 2/(6x^2+13x-5) is x 1/3

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domain of f(x)= 2/(6x^2+13x-5)

domain

f (x) = \frac{2}{6 x ^{2} + 13 x - 5}

x < - \frac{5}{2} or - \frac{5}{2} < x < \frac{1}{3} or x > \frac{1}{3}

Interval Notation

(- \infty, - \frac{5}{2}) \cup (- \frac{5}{2}, \frac{1}{3}) \cup (\frac{1}{3}, \infty)

Solution steps

Find undefined (singularity) points:

x = \frac{1}{3}, x = - \frac{5}{2}

The function domain

x < - \frac{5}{2} or - \frac{5}{2} < x < \frac{1}{3} or x > \frac{1}{3}

3 cos (4 x)

(6.8) (10.4)

\frac{- 3 x ^{2} - 12 x - 9}{x ^{2} + 5 x + 4}

49 - x^{2}

f (x) = 80 - 4.9 t^{2}

What is the domain of f(x)= 2/(6x^2+13x-5) ?
The domain of f(x)= 2/(6x^2+13x-5) is x<-5/2 \lor-5/2 <x< 1/3 \lor x> 1/3