What is the domain of f(x)=(sqrt(x^2))/(9x^2+8x-1) ?

The domain of f(x)=(sqrt(x^2))/(9x^2+8x-1) is x 1/9

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domain of f(x)=(sqrt(x^2))/(9x^2+8x-1)

domain

f (x) = \frac{x ^{2}}{9 x ^{2} + 8 x - 1}

x < - 1 or - 1 < x < \frac{1}{9} or x > \frac{1}{9}

Interval Notation

(- \infty, - 1) \cup (- 1, \frac{1}{9}) \cup (\frac{1}{9}, \infty)

Solution steps

Find non-negative values for radicals:

- \infty < x < \infty

Find undefined (singularity) points:

x = \frac{1}{9}, x = - 1

Combine real regions and undefined points for final function domain

x < - 1 or - 1 < x < \frac{1}{9} or x > \frac{1}{9}

f (x) = 100 + 15 y

7

f (x) = 3 + (8 + x)^{\frac{1}{2}}

f (x) = y + 3

What is the domain of f(x)=(sqrt(x^2))/(9x^2+8x-1) ?
The domain of f(x)=(sqrt(x^2))/(9x^2+8x-1) is x<-1\lor-1<x< 1/9 \lor x> 1/9