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

1 Answer
Apr 6, 2018

Answer:

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]#.