Question

Consider the function f defined by f(x) =x-x(x),where x is a positive variable,and (x) denotes the integral part of x and show that it is discontinuous for integral values of x,and continuous for all others. Is the function periodic? If periodic,what is its period? Draw its graph.

Answer 1

From the desfinition fo the function it follows that $f\left(x\right)=x-(n-1)forn-1<x<n....\left(1\right)$ $f\left(x\right)=0forx=n,...\left(2\right)$ $f\left(x\right)=x-nforn<x<n+1...\left(3\right)$ where n is an integer.We test the function for continuity at x=n.We have $f\left(n\right)=0.$ by (2) $f(n-0)=\underset{x\to n-0}{lim}[x-(n-1)]$ $L=\underset{h\to 0}{lim}[(n-h)-(n-1)]=1by\left(1\right)$ and $f(n+0)=\underset{x\to n+0}{lim}(x-n)$ $R=\underset{h\to 0}{lim}(n+h-n)=0,by\left(3\right)$ Hence f is discontinuous at x=n i.e. for all integral values ofx.It is obviously contin uous for all other values.Since x is positive variable, putting n=1,2,3,4,5,... We see that the graph consists of the following: $\begin{array}{ccc}y=x& when& 0<x<1\\ y=0& when& x=1\\ y=x-1& when& 1<x<2\\ y=0& when& x=2\\ y=x-2& when& 2<x<3\\ y=0& when& x=3\\ y=x-3& when& 3<x<4\\ y=0& when& x=4\end{array}$

and so on. The graph is shown by thick lines from x=0 to x=4.

Remark : From the graph the following facts are evident :

(i) The function is discountinuous for all integral values and continuous for all others.

(ii) In every range which includes an integer, it is bouned between 0 and 1

.(iii) The lower bound zero is attained but the upper bound 1 is not attained since $f\left(x\right)\ne 1$ for any value of x whatsoever.