What is the domain of f(x)=1+sqrt(x(x-3)) ?

The domain of f(x)=1+sqrt(x(x-3)) is x = 3

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domain of f(x)=1+sqrt(x(x-3))

domain

f (x) = 1 + x (x - 3)

x \leq 0 or x \geq 3

Interval Notation

(- \infty, 0] \cup [3, \infty)

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

f (x) = \frac{x ^{3} + 3 x ^{2} + 1}{4 x ( x - 1 ) ^{2}}

f (x) = \frac{1}{( y - 2 ) + ( y - 8 )}

f (x) = - \frac{3}{4} x + 2

f (x) = ∣ x ∣ + 4

\frac{x - 2}{5 - x}

What is the domain of f(x)=1+sqrt(x(x-3)) ?
The domain of f(x)=1+sqrt(x(x-3)) is x<= 0\lor x>= 3