What is the domain of h(x)= 1/(\sqrt[4]{x^2-5x)} ?

The domain of h(x)= 1/(\sqrt[4]{x^2-5x)} is x 5

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domain of h(x)= 1/(\sqrt[4]{x^2-5x)}

domain

x < 0 or x > 5

Solution steps

Find non-negative values for radicals:

x \leq 0 or x \geq 5

Find undefined (singularity) points:

x = 5, x = 0

Combine real regions and undefined points for final function domain

x < 0 or x > 5

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

- 6 (0.5)

\frac{x + 2}{x ^{2} - 1}

f (x) = 5 x

f (x) = x^{4} + 4 x^{3}

What is the domain of h(x)= 1/(\sqrt[4]{x^2-5x)} ?
The domain of h(x)= 1/(\sqrt[4]{x^2-5x)} is x<0\lor x>5