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$f(x)$ is a monic real polynomial with $f(1/x) = f''(x) = c$, when $x$ tends to infinity, and $f(x)\ge f(1)$ for all real $x$. What is $f$?
My attempt: I showed that, as $x$ tends to infinity $f(1/x) = c = f''(x)$, f has degree $\le 3$ and constant $c$. Thus let $f(x) = x^3 + ax^2 + bx + c$. Then using the last given condition I tried to find the co-efficients, but that seemed to be a long process. I wonder if there's a shorter and more elegant version of this proof.
If $f^{''}(x) \rightarrow c$ then $f$ must be a second degree polynomial. That way, $$f(x) = x^2+bx+a \Rightarrow f^{''}(x) = 2 = c$$ Furthermore: $$f\left(\frac{1}{x}\right) = \left(\frac{1}{x}\right)^2 + b\left(\frac{1}{x}\right) + a = \frac{1 + bx + ax^2}{x^2}\rightarrow a, \text{ as } x \rightarrow\infty$$ That way, $$a = 2$$
I think the a best way to guarantee that $f(x)$ have a minimum at $x=1$ is to demand that: $$f^{'}(1) = 0$$ since: $$f^{'}(x) = 2x+b$$ we have that: $$f^{'}(1) = 2+b = 0 \Rightarrow b = -2 $$ and: $$f(x) = x^2-2x+2$$
$$f(x)=x^n+a_{n-1}x^{n-1}+...+a_0 \rightarrow f''(x)=n(n-1)x^{n-2}+...+6a_3x+2a_2$$
If $f''(x) \rightarrow c$ when $x \rightarrow \infty$ then $f$ must be at most a second degree polynomial.
$1)$ If $f$ is a second degree then:
$$f(x)=x^2+bx+c \rightarrow f''(x)=2$$
It means that $\lim_{x \rightarrow \infty}f(1/x)=2$ but
$$f\left(\frac{1}{x}\right)=\left(\frac{1}{x}\right)^2+b\left(\frac{1}{x}\right)+c \rightarrow c=2$$
then
So, for now we have $f(x)=x^2+bx+2$. Now $f(x)\ge f(1)$ for all $x \in \Bbb R$ then
$$x^2+bx+2 \ge 1+b+2 \rightarrow x^2+bx \ge 1+b $$
$$x^2+bx-b-1 \ge 0 \quad \text{for all} \quad x$$
It means that
$$\Delta =b^2+4(b+1) = (b+2)^2 \le 0 \rightarrow b=-2$$
and then $f(x)=x^2-2x+2$.
$2)$ What if $f(x)=x+c$, or $f(x)=c$? Can you finish?
