1

Problem: If $f(x)=n$, where $n$ is an integer such that $n\le x\lt n+1$, what is the range of $f(x)$?

The answer in the book is the set of all integers. Since this is the answer, isn't $f(x)=n$ enough information to solve this problem? What is the point of the problem having the $n\le x\lt n+1$ part?

hardmath
  • 37,015
54284User
  • 111
  • 2
    Without the condition $n \leq x < n + 1$, you wouldn't know which $n$ you are talking about. For example, we wouldn't know that $f(2.5) = 2$. – J126 Jun 26 '18 at 02:45
  • It might help for such Questions to identify the textbook's author and title. – hardmath Jun 26 '18 at 02:53

1 Answers1

2

The point is to tell you what function $f$ refers to.

For example, it tells you that $f(4.3)$ is equal to that integer value of $n$ for which $n \leq 4.3 < n + 1$. So $f(4.3) = 4$. Similarly, $f(2.5) = 2$, $f(7) = 7$, $f(-3.9) = -4$, and so on.

If it just said $f(x) = n$, $f$ would be a constant function equal to some unspecified integer $n$. For example, if $n$ were $5$, then we'd have $f(4.3) = 5$, $f(2.5) = 5$, $f(7) = 5$, $f(-3.9) = 5$.

Dave
  • 101