What is the domain of f(x)=\sqrt[4]{x^2-3x} ?

The domain of f(x)=\sqrt[4]{x^2-3x} is x = 3

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domain of f(x)=\sqrt[4]{x^2-3x}

domain

x \leq 0 or x \geq 3

Solution steps

Find non-negative values for radicals:

x \leq 0 or x \geq 3

The function domain

x \leq 0 or x \geq 3

y = \frac{2}{3} x - 2

\frac{2 x - 2}{3 ( x ^{2} - 2 x ) ^{\frac{2}{3}}}

f (x) = \frac{x + 3}{x - 4}

y + 5 = - 4 (x + 6)

What is the domain of f(x)=\sqrt[4]{x^2-3x} ?
The domain of f(x)=\sqrt[4]{x^2-3x} is x<= 0\lor x>= 3