What is the domain of y=sqrt(x^2-x) ?

The domain of y=sqrt(x^2-x) is x = 1

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domain of y=sqrt(x^2-x)

domain

y = x^{2} - x

x \leq 0 or x \geq 1

Interval Notation

(- \infty, 0] \cup [1, \infty)

Solution steps

Find non-negative values for radicals:

x \leq 0 or x \geq 1

The function domain

x \leq 0 or x \geq 1

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

f (x) = arccos (x - 3)

3 + 16 - (x - 3)^{2}

f (x) = \frac{- 2 \cdot 0}{4} + e^{0} - 13

y = 5 x - 2

What is the domain of y=sqrt(x^2-x) ?
The domain of y=sqrt(x^2-x) is x<= 0\lor x>= 1