# How do you find the domain of f(x)=x/(6x^2+13x-5)?

Domain (value of x) is any real value except $x = \frac{1}{3} \mathmr{and} x = - \frac{2}{5}$
Domain is the value of x in the function. Here the denomintor must not be zero. $6 {x}^{2} + 13 x - 5 \ne 0 \mathmr{and} 6 {x}^{2} + 15 x - 2 x - 5 \ne 0 \mathmr{and} 3 x \left(2 x + 5\right) - 1 \left(2 x + 5\right) \ne 0 \mathmr{and} \left(3 x - 1\right) \left(2 x + 5\right) \ne 0 \mathmr{and} x \ne \frac{1}{3} \mathmr{and} x \ne - \frac{5}{2} \therefore$Domain (value of x) is any real value except $x = \frac{1}{3} \mathmr{and} x = - \frac{2}{5}$ [Ans]