If $f(x)=\sqrt{3x-2}$, what is the smallest possible value of $f(x)$ ?

On this question you can use Plugging In the Answers. The numbers in the answer choices replace the $f(x)$ portion of the equation, so you can just write out the rest of it, $\sqrt{3x-2}$, next to each to see if it can be true. Start with (A) since you are looking for the smallest value of $f(x)$. If $0=\sqrt{3x-2}$, then $0=3x-2$ when you square both sides. Add 2 to both sides to get $2=3x$, and then divide both sides by 3 . You get $x=\frac{2}{3}$. Since this is a real value, the equation works, so the smallest value of $f(x)$ is 0 . Choice (A) is correct.