Problem: Let [x] denote the largest integer (positive, negative or zero) less than or equal to x. Let y= f(x) = [x] + \sqrt{x - [x]}  be defined for all real numbers x.

(i) Sketch on plain paper, the graph of the function f(x) in the range -5 \le x \le 5
(ii) Show that, any given real number y_0  , there is a real number x_0  such that y_0 = f(x_0)


First note that \sqrt{x - [x]}  is same as \sqrt{t} , 0\le t \le 1  .

It’s graph between 0 to 1 looks like:

Screen Shot 2015-11-29 at 9.40.45 PM

Clearly [x] part only increments (or decrements) it by integer quantity as [x] is constant between any two integers. That for any integer k  for all \( x \in (k, k+1) \).
f(x) = k +\sqrt{t}  , t\in(0,1)  . Hence graph of f(x) is as follows:

Screen Shot 2015-11-29 at 9.47.31 PM

Finally consider and arbitrary value y_0  . We take x_0 = [y_0] + (y - [y_0])^2 . Then f(x_0) = [x_0] + \sqrt(x - [x_0] = [y_0] + \sqrt{(y - [y_0])^2} = y_0  (since 0 \le (y - [y_0]) < 1 \Rightarrow 0 \le (y - [y_0])^2 < 1  )


  • What is this topic: Graphing of functions
  • What are some of the associated concept: Greatest Integer Function
  • Where can learn these topics: Cheenta I.S.I. & C.M.I. course, discusses these topics in the ‘Calculus’ module.
  • Book Suggestions: Play with Graphs, Arihant Publication