What is the domain of f(x)= 1/((3x^2+17x-6)) ?

The domain of f(x)= 1/((3x^2+17x-6)) is x 1/3

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domain of f(x)= 1/((3x^2+17x-6))

domain

f (x) = \frac{1}{( 3 x ^{2} + 17 x - 6 )}

x < - 6 or - 6 < x < \frac{1}{3} or x > \frac{1}{3}

Interval Notation

(- \infty, - 6) \cup (- 6, \frac{1}{3}) \cup (\frac{1}{3}, \infty)

Solution steps

Find undefined (singularity) points:

x = \frac{1}{3}, x = - 6

The function domain

x < - 6 or - 6 < x < \frac{1}{3} or x > \frac{1}{3}

f (x) = ∣ x ∣ + 1

f (x) = x - 2 - 1

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

f (x) = arcsin (ln (x))

\frac{2 x + 1}{x ^{2} + 3 x - 10}

What is the domain of f(x)= 1/((3x^2+17x-6)) ?
The domain of f(x)= 1/((3x^2+17x-6)) is x<-6\lor-6<x< 1/3 \lor x> 1/3