$p(x)$ is a polynomial in $R[x]$ such that $p(0)=1$ , $p(x) \ge p(1)$ and $\lim_{x \rightarrow \infty} p''(x)=4$ Find $p(2)$ $p(x)$ is a polynomial in $R[x]$ such that $p(0)=1$ , $p(x) \ge p(1)$ and $\lim_{x \rightarrow \infty} p''(x)=4$ Find $p(2)$
I assumed the degree of $P(x)$ to be smaller ones like $2,3$ and found $P(2)$ to be $1$
But I dont know how to generalise.
 A: Suppose that the degree of $p''(x)$ is equal to or larger than $1$. 
Let $$p''(x):=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_1x+a_0$$
where $a_n\not=0$ and $n\ge 1$.
Then,
$$\lim_{x\to\infty}p''(x)=\lim_{x\to\infty}x^n(a_n+a_{n-1}x^{-1}+a_{n-2}x^{-2}+\cdots +a_1x^{1-n}+a_0x^{-n})$$
This is equal to $+\infty$ if $a_n\gt 0$, or is equal to $-\infty$ if $a_n\lt 0$. 
This contradicts that $\lim_{x\to\infty}p''(x)=4$.
It follows from this that $p''(x)=4$ and that the degree of $p(x)$ is $2$.
We can set $p(x)$ as $p(x)=2x^2+bx+1$.
We have 
$$2x^2+bx+1\ge 2+b+1,$$
i.e.
$$2x^2+bx-2-b\ge 0$$
So, we have to have
$$b^2-4\cdot 2(-2-b)\le 0\iff (b+4)^2\le 0\iff b=-4$$
Therefore, $p(2)=\color{red}{1}$.
A: The polynomial has degree $2$, because if it has degree less than $2$ its second derivative is zero and if it has degree greater than $2$ its second derivative has no finite limit at $\infty$, being a polynomial of degree greater than $1$.
Thus $p(x)=ax^2+bx+c$ and $p''(x)=2a$. Therefore $a=2$.
Since the graph is a parabola, we know from $p(x)\ge p(1)$ that $1$ is the $x$-coordinate of the vertex, so
$$
1=-\frac{b}{2a}=-\frac{b}{4}
$$
and therefore $b=-4$. Since $c=p(0)=1$, the polynomial is
$$
p(x)=2x^2-4x+1
$$
and $p(2)=1$.
