1. find 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?

2. A 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.

3. Originally Posted by TwistedOne151
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.
i think f(x) has x-1 x-2 x-3
so i assume f(x)=a(x-1)(x-2)(x-3) is it right?
but
i can not slove the problem~

4. 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.