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]} ,s=2 $ be defined for all real numbers x.
(i) Sketch on plain paper, the graph of the function f(x) in the range $-5 \le x \le 5 ,s=2$
(ii) Show that, any given real number $y_0 ,s=2 $, there is a real number $x_0 ,s=2 $ such that $y_0 = f(x_0) ,s=2 $
Discussion:
First note that $\sqrt{x - [x]} ,s=2 $ is same as $\sqrt{t} , 0\le t \le 1 ,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} ,s=2 $ , $t\in(0,1) ,s=2 $. Hence graph of f(x) is as follows:
Finally consider and arbitrary value $ y_0 ,s=2 $. We take $x_0 = [y_0] + (y - [y_0])^2 ,s=2$. Then $f(x_0) = [x_0] + \sqrt(x - [x_0] = [y_0] + \sqrt{(y - [y_0])^2} = y_0 ,s=2 $ (since $0 \le (y - [y_0]) < 1 \Rightarrow 0 \le (y - [y_0])^2 < 1 ,s=2 $ )
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]} ,s=2 $ be defined for all real numbers x.
(i) Sketch on plain paper, the graph of the function f(x) in the range $-5 \le x \le 5 ,s=2$
(ii) Show that, any given real number $y_0 ,s=2 $, there is a real number $x_0 ,s=2 $ such that $y_0 = f(x_0) ,s=2 $
Discussion:
First note that $\sqrt{x - [x]} ,s=2 $ is same as $\sqrt{t} , 0\le t \le 1 ,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} ,s=2 $ , $t\in(0,1) ,s=2 $. Hence graph of f(x) is as follows:
Finally consider and arbitrary value $ y_0 ,s=2 $. We take $x_0 = [y_0] + (y - [y_0])^2 ,s=2$. Then $f(x_0) = [x_0] + \sqrt(x - [x_0] = [y_0] + \sqrt{(y - [y_0])^2} = y_0 ,s=2 $ (since $0 \le (y - [y_0]) < 1 \Rightarrow 0 \le (y - [y_0])^2 < 1 ,s=2 $ )
