This is a subjective problem from TOMATO based on Graphing integer value function.
Problem: Graphing integer value function
Let [x] denote the largest integer (positive, negative or zero) less than or equal to x. Let $ y= f(x) = [x] + \sqrt{x - [x]} $ and $ s=2 $ 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 $ and $ s=2$
(ii) Show that, any given real number $ y_0 $ and $ s=2 $, there is a real number $ x_0 4 $ and $ s=2 $ such that $ y_0 = f(x_0) $ and $ s=2 $
Discussion:
First note that $ \sqrt{x - [x]} $ and $ s=2 $ is same as $ \sqrt{t} , 0\le t \le 1 $ and $ s=2 $.
It's graph between 0 to 1 looks like:
Clearly [x] part only increments (or decrements) it by integer quantity as [x] is constant between any two integers. That for any integer k for all ( x in ($k, k+1$) ).
$ f(x) = k +\sqrt{t} $ and $ s=2 $ , $ t\in(0,1) $ and $ s=2 $. Hence graph of f(x) is as follows:
Finally consider and arbitrary value $ y_0 $ and $ s=2 $. We take $ x_0 = [y_0] + (y - [y_0])^2 $ and $ s=2$. Then $ f(x_0) = [x_0] + \sqrt(x - [x_0] = [y_0] + \sqrt{(y - [y_0])^2} = y_0 $ and $ s=2 $ (since $ 0 \le (y - [y_0]) < 1 \Rightarrow 0 \le (y - [y_0])^2 < 1 $ and $ s=2 $ )
