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

The domain of (sqrt(x))/(9x^2+8x-1) is 0 1/9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

domain

\frac{x}{9 x ^{2} + 8 x - 1}

0 \leq x < \frac{1}{9} or x > \frac{1}{9}

Interval Notation

[0, \frac{1}{9}) \cup (\frac{1}{9}, \infty)

Solution steps

Find non-negative values for radicals:

x \geq 0

Find undefined (singularity) points:

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

Combine real regions and undefined points for final function domain

0 \leq x < \frac{1}{9} or x > \frac{1}{9}

f (x) = \frac{x - 12}{x + 19}

x^{2} - 4 x - 12

\frac{2}{x}

(- 6, - 1), (5, - 5)

f (x) = - 2 x - 3 + 1

What is the domain of (sqrt(x))/(9x^2+8x-1) ?
The domain of (sqrt(x))/(9x^2+8x-1) is 0<= x< 1/9 \lor x> 1/9