find the polynomial ft f(x) with the smallest degree so that

$\displaystyle \lim_{x->1}\frac{f(x)}{x-1}\$=1

and

$\displaystyle \lim_{x->2}\frac{f(x)}{x-2}\$=2

and

$\displaystyle \lim_{x->3}\frac{f(x)}{x-3}\$=3

pls how should i assume the ft?

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- Feb 14th 2010, 10:09 PMchialin4find the polynomial problem
find the polynomial ft f(x) with the smallest degree so that

$\displaystyle \lim_{x->1}\frac{f(x)}{x-1}\$=1

and

$\displaystyle \lim_{x->2}\frac{f(x)}{x-2}\$=2

and

$\displaystyle \lim_{x->3}\frac{f(x)}{x-3}\$=3

pls how should i assume the ft? - Feb 14th 2010, 11:42 PMTwistedOne151A hint
Consider what the existence of those limits (and their finite, non-zero values) means for the values of f(x) at x=1, x=2, and x=3. What monomials must then be factors of the polynomial f(x)?

--Kevin C. - Feb 15th 2010, 12:10 AMchialin4
- Feb 15th 2010, 12:21 AMTwistedOne151
Chialin4,

Close. f(x)=(x-1)(x-2)(x-3)*p(x), where p(x) is a polynomial with nonzero values at x=1, x=2, and x=3. Consider what the limits say about p(x), and find the lowest degree polynomial that satisfies these conditions.

--Kevin C.