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

The domain of \sqrt[4]{x^2-4x} is x = 4

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domain of \sqrt[4]{x^2-4x}

domain

4 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

f (x) = 36 - 12 x

\frac{9 + x}{1 - 9 x}

f (x) = \frac{4 + 7 x}{x - 1}

f (x) = lo g_{10} (9 x^{2} + 30 x + 29)

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

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