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

The domain of y=2+sqrt(x^2-4x) is x = 4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

domain of y=2+sqrt(x^2-4x)

domain

y = 2 + x^{2} - 4 x

x \leq 0 or x \geq 4

Interval Notation

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

Solution steps

Find non-negative values for radicals:

x \leq 0 or x \geq 4

The function domain

x \leq 0 or x \geq 4

121 - x^{2}

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

\frac{t - 5}{t ^{2} + 2 t - 15}

f (x) = 2 x + 5, 0 \leq x \leq 2

- 2 x^{2} + 2 x

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