What is the domain of f(x)=\sqrt[4]{x^2+3x} ?

The domain of f(x)=\sqrt[4]{x^2+3x} is x = 0

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domain of f(x)=\sqrt[4]{x^2+3x}

domain

x \leq - 3 or x \geq 0

Interval Notation

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

Solution steps

Find non-negative values for radicals:

x \leq - 3 or x \geq 0

The function domain

x \leq - 3 or x \geq 0

x^{4} - 1

f (x) = \frac{4}{x + 1}

f (x) = ln (e^{x + 1}) + ln (2) - 1

(8.2) (3.8)

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

What is the domain of f(x)=\sqrt[4]{x^2+3x} ?
The domain of f(x)=\sqrt[4]{x^2+3x} is x<=-3\lor x>= 0