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) = 6 (2)^{x - 3} + 7

f (x) = \frac{1}{18 - 3 x - x ^{2}}

f (x) = \frac{x ^{2} - 4}{x}

f (x) = - lo g_{9} (x + 5)

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