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

1 Answer
Apr 6, 2018

Domain: [17, infty)[17,)

Explanation:

One cannot have a negative under a square root, so we know 17 - x >= 017x0. Adding xx to both sides yields 17 >= x17x. Thus, xx can be any number greater than or equal to 1717. This gives the interval [17, infty)[17,) as our domain.

To elaborate, sqrt(n)n asks, "what number, when squared, gives nn". Notice that positive numbers, when squared, give positive numbers. (2^2 = 422=4) Also, negative numbers, when squared, give positive numbers. (-2^2 = (-2)(-2) = 422=(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 - x17x must be non-negative. This is written as the inequality 17 - x >= 017x0. Algebraic manipulation gives 17 >= x17x, and from this we extrapolate our interval [17, infty][17,].