What is the domain of f(x)=arcsin(((x^2-4))/(x^2+1)) ?

The domain of f(x)=arcsin(((x^2-4))/(x^2+1)) is x = sqrt(3/2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

domain of f(x)=arcsin(((x^2-4))/(x^2+1))

domain

f (x) = arcsin (\frac{( x ^{2} - 4 )}{x ^{2} + 1})

x \leq - \frac{3}{2} or x \geq \frac{3}{2}

Interval Notation

(- \infty, - \frac{3}{2}] \cup [\frac{3}{2}, \infty)

Solution steps

Find known functions domain restrictions:

x \leq - \frac{3}{2} or x \geq \frac{3}{2}

The function domain

x \leq - \frac{3}{2} or x \geq \frac{3}{2}

\frac{x + 3}{- 2 x + 8}

e^{- y} + e^{2}

\frac{x + 9}{x ^{2} - 81}

f (x) = \frac{x ^{2}}{x ^{2} - x - 30}

f (x) = - 5 x + 1

What is the domain of f(x)=arcsin(((x^2-4))/(x^2+1)) ?
The domain of f(x)=arcsin(((x^2-4))/(x^2+1)) is x<=-sqrt(3/2)\lor x>= sqrt(3/2)