What is the domain of f(x)=2+sqrt(x^2-4x) ?

The domain of f(x)=2+sqrt(x^2-4x) is x = 4

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domain of f(x)=2+sqrt(x^2-4x)

domain

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

x \leq 0 or x \geq 4

Interval Notation

(- \infty, 0] \cup [4, \infty)

Solution steps

Find non-negative values for radicals:

x \leq 0 or x \geq 4

The function domain

x \leq 0 or x \geq 4

f (x) = lo g_{2} (\frac{3 x - 2}{x - 1} + 1)

y = lo g_{10} (5 + 3 x)

f (x) = (2 x)^{\frac{1}{3}} - 7

f (x) = \frac{5}{2 x + 1}

f (x) = 2 π r^{2} + 12 π r

What is the domain of f(x)=2+sqrt(x^2-4x) ?
The domain of f(x)=2+sqrt(x^2-4x) is x<= 0\lor x>= 4