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

The domain of f(x)=4+sqrt(x^2+4x) is x = 0

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

domain

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

x \leq - 4 or x \geq 0

Interval Notation

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

Solution steps

Find non-negative values for radicals:

x \leq - 4 or x \geq 0

The function domain

x \leq - 4 or x \geq 0

f (x) = 4 - x

f (x) = 150 x - \frac{3 x ^{2}}{2}

f (x) = \frac{1}{12 x + 17}

2 ∣ x ∣ + 3 - x^{2}

f (x) = x y - 2 y - x = 0

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