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

The domain of f(x)=sqrt(x^2+4x) 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)=sqrt(x^2+4x)

domain

f (x) = x^{2} + 4 x

x \leq - 4 or x \geq 0

Interval Notation

(- \infty, - 4] \cup [0, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - 4 or x \geq 0

The function domain

x \leq - 4 or x \geq 0

f (x) = \frac{2 x - 7}{3}

\frac{2}{3} x - 12

e^{2 x} - 3

f (x) = ∣ 8 + 6 x ∣

y = \frac{x ^{2} + x}{x + 1}

What is the domain of f(x)=sqrt(x^2+4x) ?
The domain of f(x)=sqrt(x^2+4x) is x<=-4\lor x>= 0