What is the domain of f(x)=sqrt(9x^2+6x) ?

The domain of f(x)=sqrt(9x^2+6x) is x = 0

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domain of f(x)=sqrt(9x^2+6x)

domain

f (x) = 9 x^{2} + 6 x

x \leq - \frac{2}{3} or x \geq 0

Interval Notation

(- \infty, - \frac{2}{3}] \cup [0, \infty)

Solution steps

Find non-negative values for radicals:

x \leq - \frac{2}{3} or x \geq 0

The function domain

x \leq - \frac{2}{3} or x \geq 0

f (x) = - 2^{5 - x} + 4

\frac{1}{1 - sin ( x )}

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

- x^{2} + 12 x - 34

f (x) = \frac{x}{3 x ^{2} + 2 x - 8}

What is the domain of f(x)=sqrt(9x^2+6x) ?
The domain of f(x)=sqrt(9x^2+6x) is x<=-2/3 \lor x>= 0