Is this function a function of bounded variation? I am relevantly new to this concept of Bounded variation.
I want to know is the function  $\sqrt {1-x^2} $, $x\in (-1,1) $ of bounded variation? How should I proceed. Is there any graphical interpretation of functions of bounded variation?
 A: As for the specific question, $f(x)=\sqrt{1-x^2}$ it is a function of bounded variation over the closed interval $[-1,1]$ (I don't know why the restriction here to the open interval, but if taken as a limit, the variation over $(-1,1)$ would also be finite, hence of bounded variation.
A simple proof relies on the following characterization:

$f$ is of bounded variation over $[a,b]$ $\iff$ there exist non decreasing functions $g,h$ over $[a,b]$ such that $f=g-h$. (*)

So, consider
$$g(x)=\left\{\begin{array}{ccc}\sqrt{1-x^2}&\text{if}&x<0\\1&\text{if}&x\ge0\\\end{array}\right.$$
and
$$h(x)=\left\{\begin{array}{ccc}0&\text{if}&x<0\\1-\sqrt{1-x^2}&\text{if}&x\ge0,\\\end{array}\right.$$
which happen to be non decreasing and verify the condition $f=g-h$.
As for a graphical interpretation, I find easier to say what a function of unbounded variation is: it's graph is such that if you count the vertical 'distance traveled' going from $a$ to $b$ (or backwards), you can say that you moved over any finite number. For instance, with the function
$$f(x)=\sin \left(\frac1x\right)$$
(for $x\neq0$ and $0$ for $x=0$; [graph])1 you can move leftwards from $x_1=\frac2{\pi}$, where $f$ value is $1$, to $x_2=\frac2{3\pi}$ while the function value will go down monotonously from $1$ to $-1$: you will have gone $2$ units down.
Now if we go on leftwards to $x_3=\frac2{5\pi}$ while $f$ goes up until reaching again the value $1$. You have moved $2$ units up now, and this means you traveled a total of $4$ units vertically: that is, the total variation of $f$ from $x_1$ to $x_3$ is 4.
Now, since taking $x_n=\frac2{(2\cdot n-1)\pi}$ gives an infinite sequence of points $0<\dots<x_3<x_2<x_1$ such that the variation between to consecutive points is always $2$, you can make the variation as large as you want considering enough many points (or, in the language used for the definition, taking a partition that is fine ('thin') enough.

ONE MORE COMMENT: The thing with this example you are asking about, is that you cannot use the following usual result:

If $f$ has bounded continuous derivative over an interval, then it is of bounded variation over that interval.

But your function has unbounded derivative over $(-1,1)$, so that's why we needed to use other ways to prove bounded variation.

(*) The proof of this property actually gives a procedure for finding such $f$ and $g$.
First of all, let's remember that the total variation of $f$ over $[a,b]$ is defined as
$$V_a^b(f)=\sup \left\{\sum_{i=1}^n |f(x_i)-f(x_{i-1})|\colon \{x_0,\ldots,x_n\}\in\mathscr P\right\},$$
where $\mathscr P$ is the set of all possible partitions of $[a,b]$. So $f$ is said to be of bounded variation over $[a,b]$ iff $V_a^b(f)<\infty$.
The central part of the proof (which I omit although is not difficult) is to see that the function $V\colon [a,b]\rightarrow \mathbb R$
$$V(x)=V_a^x(f)$$
is non-decreasing.
Then, you can take $g(x)=V(x)$ and $h(x)=V(x)-f(x)$, which can also be proven to be non-decreasing. That of course implies $g(x)-h(x)=f(x)$.
In this example I saw that $f(x)$—being increasing from $-1$ to $0$—gave itself the total variation over this interval: think that in that case $|f(x_i)-f(x_{i-1})|=f(x_i)-f(x_{i-1})$, and so 
$$\sum_{i=1}^n |f(x_i)-f(x_{i-1})|=f(x_1)-f(x_0)+f(x_2)-f(x_1)+$$
$$+f(x_3)-f(x_2)+\cdots+f(x_{n-1})-f(x_{n-2})+f(x_n)-f(x_{n-1})=f(x_n)-f(x_0).$$
That is, for $x<0$, $V(x)=f(x)-f(-1)=f(x).$
Then, from $0$ to $1$ $f$ was decreasing, so $V_0^x(f)=f(0)-f(x)$. I could have defined here then
$$g(x)=V(x)=\left\{\begin{array}{ccc}\sqrt{1-x^2}&if&x<0\\1-\sqrt{1-x^2}&if&x\ge0,\\
\end{array}\right.$$
and so $h$ would have been
$$h(x)=g(x)-f(x)=\left\{\begin{array}{ccc}0&if&x<0\\1-2\sqrt{1-x^2}&if&x\ge0,\\\end{array}\right.$$
Of course, this is not the only way, since you can see that what I did above is different. I haven't think that much at the moment, but let's say that I found more natural to take account of the 'increasing' variations in $g$ and of the 'decreasing' variations in $h$, and that's why I kept constant $g$ after $x=0$. This had the effect of making $h=g-f$ constant when $f$ was increasing and take account of the variation when $f$ was decreasing. Just two of many ways to pick $g$ and $h$ when using this property.
As for the reciprocal ($f=g-h$ with $g,h$ non-decreasing implies $f$ is of bounded variation) is immediate from the observation that monotone functions are of bounded variation and from the fact that $V_a^b(cf)=|c|V_a^b(f)$ and $V_a^b(f+g)\le V_a^b(f)+V_a^b(g)$.
