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

The domain of f(x)=(2x)/(sqrt(2x-1)-1) is 1/2 1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

domain of f(x)=(2x)/(sqrt(2x-1)-1)

domain

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

\frac{1}{2} \leq x < 1 or x > 1

Interval Notation

[\frac{1}{2}, 1) \cup (1, \infty)

Solution steps

Find non-negative values for radicals:

x \geq \frac{1}{2}

Find undefined (singularity) points:

x = 1

Combine real regions and undefined points for final function domain

\frac{1}{2} \leq x < 1 or x > 1

r an g e \frac{1}{9 - x ^{2}}

d o main \frac{1}{x + 6}

e x t re m e (x - 2) (x + 4) (x - 6) (x + 8)

d o main f (x) = \frac{x + 7}{2 - x}

a sy m pt o t es f (x) = \frac{( x ^{2} - x )}{x ^{2} - 9 x + 8}

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