What is the domain of \sqrt[4]{x^2-7x} ?

The domain of \sqrt[4]{x^2-7x} is x = 7

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domain of \sqrt[4]{x^2-7x}

domain

x \leq 0 or x \geq 7

Interval Notation

(- \infty, 0] \cup [7, \infty)

Solution steps

Find non-negative values for radicals:

x \leq 0 or x \geq 7

The function domain

x \leq 0 or x \geq 7

f (x) = (x^{3} - 1)

f (x) = x^{2} - 8 x + 15

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

f (x) = - e^{- x} + 6

f (x) = - e^{- x} + 8

What is the domain of \sqrt[4]{x^2-7x} ?
The domain of \sqrt[4]{x^2-7x} is x<= 0\lor x>= 7