# What is the domain of  f(x)=sqrt(17-x)?

Domain: $\left[17 , \infty\right)$
One cannot have a negative under a square root, so we know $17 - x \ge 0$. Adding $x$ to both sides yields $17 \ge x$. Thus, $x$ can be any number greater than or equal to $17$. This gives the interval $\left[17 , \infty\right)$ as our domain.
To elaborate, $\sqrt{n}$ asks, "what number, when squared, gives $n$". Notice that positive numbers, when squared, give positive numbers. (${2}^{2} = 4$) Also, negative numbers, when squared, give positive numbers. ($- {2}^{2} = \left(- 2\right) \left(- 2\right) = 4$) So it follows that one cannot take the square root of a negative number, since no number, when squared, yields another negative number.
When we realize that, we know that $17 - x$ must be non-negative. This is written as the inequality $17 - x \ge 0$. Algebraic manipulation gives $17 \ge x$, and from this we extrapolate our interval $\left[17 , \infty\right]$.