#### Answer

$\left[ \frac{-5}{3},\infty\right)$

#### Work Step by Step

We want to determine all real $x$ values for which $f(x)=(3x+5)^{1/2}$ is defined. We can rewrite $f(x)$ as
$$f(x)=\sqrt{3x+5}.$$
We can only take the square root of a nonnegative real number, so set the inside of the square root greater than or equal to 0 to find all $x$ where $f$ is defined:
$$3x+5 \geq 0.$$
Subtracting 5 from both sides of the inequality gives:
$$3x \geq -5$$
Divide both sides by 3 to obtain:
$$x \geq \frac{-5}{3}$$.
Therefore, the domain of $f$ is $x \geq \frac{-5}{3}$, which can be written as $\left[ \frac{-5}{3},\infty\right)$ in interval notation.