Let \(f(x)\) be a polynomial of degree \(n\) such that \(f(k)=\frac{k}{k+1}\) for \(k=0,1,2,...,n\). Find \(f(n+1)\).

Help would be greatly appreciated.

Victor

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## Comments

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TopNewestLet \(g(x)\) be the polynomial \[g(x) = (x + 1) f(x) - x.\] What can you say about \(g(x)\)? (Think roots and degree.)

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Comment deleted Apr 20, 2014

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I agree since it is defined it doyle be kequals n+1/n+2

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No, f(k) is only true for k=1,2,...,n. No one said it was true for k=n+1

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