What is the domain of f(x)=\sqrt[4]{x^2-4x} ?

The domain of f(x)=\sqrt[4]{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 f(x)=\sqrt[4]{x^2-4x}

domain

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

11 x + 4 y = 1

f (x) = cos (2) x

f (x) = \frac{30 x ^{2}}{( 6 - 5 x ^{3} ) ^{3}}

y = 2 x^{3} - 12 x^{2} + 10 x + 10

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

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