Find $p(x) = ax^3 + bx^2 + cx + d $, where a, b, c and d are real constants, satisfying the following conditions:
$1)\quad 4x^3 - 12x^2 + 12x - 3 \le p(x) \le 2019(x^3 - 3x^2 + 3x) - 2018,$ for all values of $x \ge 1$.
$2)\quad p(2) = 2011.$
My idea: The condition $(1) \iff 4(x - 1)^3 + 1\le p(x) \le 2019(x - 1)^3 + 1 $ , for all $x \ge 1.$
Therefore, I predict $p(x) = a(x - 1)^3 + 1 $, where $4 \le a \le 2019 $ . But, I still don't prove this. Please everyone help me.
Continue with what you obtained,
$$4(x - 1)^3 + 1\le p(x) \le 2019(x - 1)^3 + 1 $$
and rearrange,
$$4(x - 1)^3 \le p(x) - 1 \le 2019(x - 1)^3 $$
Assume $p(x) - 1 = a(x-1)^3$ and substitute $p(2) = 2011$ to obtain $a = 2010$. Thus,
$$p(x) = 2010(x-1)^3+1$$
Edit:
Assume, instead, a general form $p(x)-1=\sum_{k=0}^{\infty}a_k(x-1)^k$ with $a_k\ge 0$. Note that,
$$\lim_{x\to\infty} a_k(x-1)^k > 2019(x-1)^3,\>\>\> k>3$$
$$\lim_{x\to\infty} a_k(x-1)^k < 4(x-1)^3,\>\>\> k<3$$
which leads to $a_k = 0 $, for $k\ne 3$.
