What is the domain of y=\sqrt[4]{x^2-9x} ?

The domain of y=\sqrt[4]{x^2-9x} is x = 9

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domain of y=\sqrt[4]{x^2-9x}

domain

y = 4 x^{2} - 9 x

x \leq 0 or x \geq 9

Interval Notation

(- \infty, 0] \cup [9, \infty)

Solution steps

Find non-negative values for radicals:

x \leq 0 or x \geq 9

The function domain

x \leq 0 or x \geq 9

f (x) = \frac{2 x - 5}{x ^{2} - 4}

f (x) = x^{3} - 2 x

f (x) = 4 (2 x - 4) - 5

\frac{1}{X ^{2} - 4}

f (x) = - 3 + e^{- x}

What is the domain of y=\sqrt[4]{x^2-9x} ?
The domain of y=\sqrt[4]{x^2-9x} is x<= 0\lor x>= 9