1f'(x)=\frac{(x-a)(1)+(x-b)(1)}{\left ( c-a \right )\left ( c-b \right )}+\frac{(x-b)(1)+(x-c)(1)}{(a-b)(a-c)} +\frac{(x-c)(1)+(x-a)(1)}{(b-c)(b-a)}
\texttt{which simplifies to zero }
Hence
\boxed{\boxed{\frac{d}{dx }\left ( f(x) \right )=0}}
$If $\mathbf{\color{red}f(x)=\frac{(x-a).(x-b)}{(c-a)(c-b)}+\frac{(x-b).(x-c)}{(a-b)(a-c)}+\frac{(x-c).(x-a)}{(b-c)(b-a)}}$\\\\\\ Then find $\mathbf{\color{green}\frac{d}{dx}\left(f(x)\right)=}$
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4 Answers
1
21f(a)= f(b)= f(c) = 1 (since f is 2 degree if it takes same value at more than two points it must be constant.)
df/dx = 0
11nasiko's post is actually modeled on http://www.targetiit.com/iit-jee-forum/posts/29-08-09-another-quadratic-equation-question-10840.html
