Range of a function Find the range of  

$$f(x)=\frac{(x-a)(x-b)}{(c-a)(c-b)}+\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}$$ where $a, b, c$ are distinct real numbers such $a\neq b\neq c\neq a$.

 A: It is equal to $1$.
Actually this polynomial is the the approximate polynomial you get in  Lagrangian Interpolation for a curve at three data points a,b,c taking value $1$ and so $f(x)$ is identically equal to $1$ as this is $2nd$ order interpolation and it takes $1$ at all three points.
A: Observe that $g(x) = f(x) - 1$ is a polynomial of degree $2$.
But $g(a) = g(b) = g(c) = 0$. Thus a quadratic has three distinct roots. This would imply that $g(x)$ is identically $0$. Therefore $f(x)=1$. And the range, therefore, is the singleton $\{1\}$.
A: This is an identity.Range of f(x) is 1.
This is because f(x)=1 is a quadratic equation.However f(a),f(b) and f(c)=1.
Since a quadratic equation has at most 2 distinct roots f(x)=1 must be an identity (i.e valid for all values of x).
A: $$f(x)=\frac{(x-a)(x-b)}{(c-a)(c-b)}+\frac{(x-b)(x-c)}{(a-b)(a-c)}+\frac{(x-c)(x-a)}{(b-c)(b-a)}$$
$$f(a)= \dfrac{(a-b)(a-c)}{(a-b)(a-c)}=1$$
$$f(b)= \frac{(b-c)(b-a)}{(b-c)(b-a)}=1$$
$$f(c)= \frac{(c-a)(c-b)}{(c-a)(c-b)}=1$$
Now what is $f'(x)?$, $f'(x)=\dfrac{2x-(a+b)}{(c-a)(c-b)}+\dfrac{2x-(b+c)}{(a-b)(a-c)}+\dfrac{2x-(c+a)}{(b-c)(b-a)}$. Now watch the house of cards fall. Then you have a constant function, and you are aware of that constant. 
