What is the domain of f(x)=sqrt(4x^2-9) ?

The domain of f(x)=sqrt(4x^2-9) is x = 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)=sqrt(4x^2-9)

domain

f (x) = 4 x^{2} - 9

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 non-negative values for radicals:

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{4 x ^{2}}{x ^{2} - 25}

y = ln (2 x^{2} - 5 x + 8)

\frac{x + 7}{x - 9}

f (x) = \frac{8}{x ^{2} - 4 x}

sin (x^{10000}) - 10000 - x^{2}

What is the domain of f(x)=sqrt(4x^2-9) ?
The domain of f(x)=sqrt(4x^2-9) is x<=-3/2 \lor x>= 3/2