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

1 Answer
Apr 6, 2018

Domain: [17, infty)

Explanation:

One cannot have a negative under a square root, so we know 17 - x >= 0. Adding x to both sides yields 17 >= x. Thus, x can be any number greater than or equal to 17. This gives the interval [17, infty) 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 = (-2)(-2) = 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 >= 0. Algebraic manipulation gives 17 >= x, and from this we extrapolate our interval [17, infty].