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

The domain of f(x)=sqrt(-x^2+x) is 0<= x<= 1

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domain of f(x)=sqrt(-x^2+x)

domain

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

0 \leq x \leq 1

Interval Notation

[0, 1]

Solution steps

Find non-negative values for radicals:

0 \leq x \leq 1

The function domain

0 \leq x \leq 1

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

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

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

f (x) = \frac{2}{3} x - 12, - 6 \leq x \leq 4

y = 2 x

What is the domain of f(x)=sqrt(-x^2+x) ?
The domain of f(x)=sqrt(-x^2+x) is 0<= x<= 1