What is the domain of f(x)=3+sqrt(4+9x) ?

The domain of f(x)=3+sqrt(4+9x) is x>=-4/9

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domain of f(x)=3+sqrt(4+9x)

domain

f (x) = 3 + 4 + 9 x

x \geq - \frac{4}{9}

Interval Notation

[- \frac{4}{9}, \infty)

Solution steps

Find non-negative values for radicals:

x \geq - \frac{4}{9}

The function domain

x \geq - \frac{4}{9}

\frac{1 + x}{1 - x}

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

\frac{x ^{2} - 3 x}{x ^{2}}

f (x) = x - \frac{5}{3}

f (x) = \frac{3 x - 5}{x - 2}

What is the domain of f(x)=3+sqrt(4+9x) ?
The domain of f(x)=3+sqrt(4+9x) is x>=-4/9