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

The domain of f(x)=(1/(sqrt(x)))^2-9 is x>0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

domain

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

x > 0

Interval Notation

(0, \infty)

Solution steps

Find non-negative values for radicals:

x \geq 0

Find undefined (singularity) points:

x = 0

Combine real regions and undefined points for final function domain

x > 0

f (x) = sin (2 x)

(- 8, - 3), (10, 9)

f (x) = x^{2} - 12 x + 1

1 + x^{2} sin^{2} (x) + x^{2} cos^{2} (x)

f (x) = - x^{2} + 2 x - 5

What is the domain of f(x)=(1/(sqrt(x)))^2-9 ?
The domain of f(x)=(1/(sqrt(x)))^2-9 is x>0