What is the domain of 1/(\sqrt[4]{x^2-9x)} ?

The domain of 1/(\sqrt[4]{x^2-9x)} is x 9

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domain of 1/(\sqrt[4]{x^2-9x)}

domain

x < 0 or x > 9

Interval Notation

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

Solution steps

Find non-negative values for radicals:

x \leq 0 or x \geq 9

Find undefined (singularity) points:

x = 9, x = 0

Combine real regions and undefined points for final function domain

x < 0 or x > 9

f (x) = \frac{x}{x - 6}

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

f (x) = \frac{1}{3 + e ^{x}}

f (x) = x - 1

(- \frac{1}{3}, \frac{2}{3}), (- \frac{11}{3}, - \frac{8}{3})

What is the domain of 1/(\sqrt[4]{x^2-9x)} ?
The domain of 1/(\sqrt[4]{x^2-9x)} is x<0\lor x>9