What is the domain of ((y-1)/(2sqrt(y)))^2 ?

The domain of ((y-1)/(2sqrt(y)))^2 is y>0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

domain of ((y-1)/(2sqrt(y)))^2

domain

(\frac{y - 1}{2 y})^{2}

y > 0

Interval Notation

(0, \infty)

Solution steps

Find non-negative values for radicals:

y \geq 0

Find undefined (singularity) points:

y = 0

Combine real regions and undefined points for final function domain

y > 0

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

\frac{3 x ^{2} - 3}{2 x ^{2}}

f (x) = x + 5 - 5 - x

f (x) = \frac{x - 9}{x}

f (x) = 2 \cdot π \cdot x^{2} + 12 \cdot π \cdot x

What is the domain of ((y-1)/(2sqrt(y)))^2 ?
The domain of ((y-1)/(2sqrt(y)))^2 is y>0