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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 $y_0 = f(x_0)$

Discussion:

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:

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:

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 $0 \le (y - [y_0]) < 1 \Rightarrow 0 \le (y - [y_0])^2 < 1$ )

## Chatuspathi:

• 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