# How to find the polynomial?

$f(x)$ 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.

• You need to have an even-degree polynomial if $f$ is to have a global minimum at $x = 1$. Have you tried a second-degree polynomial? Jan 8, 2017 at 13:04
• What do you mean by "$f(1/x)= f''(x)=c$ when $x$ tends to infinity"? Do you mean that $\lim_{x \to \infty} f(1/x) = \lim_{x \to \infty} f''(x) =c$? Jan 8, 2017 at 13:04

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?