Polynomial such that $f''(x) \rightarrow2$ as $x\rightarrow\infty$ given some values what is $f(1)$?

Question

Let $f$ be a polynomial such that $f''(x) \rightarrow2$ as $x\rightarrow\infty$, the minimum of f is attained at $3$, and $f(0)=3$, Then $f(1)$ equals.

$(A) \ 1$

$(B) \ 2$

$(C) -1$

$(D) -2$

I am not sure how to deal with $f''(x) \rightarrow2$ as $x\rightarrow\infty$ . Give me a hint to try.

EDIT : Work after hint

Let $f(x)=ax^2+bx +c $

$f''(x)=2\implies 2=2a \implies a=1 $

$f(0)=3=c$

$f(x)=x^2+bx +3 $

Using the fact that minima attained at 3 we have $f(3)=12+3b=3 \implies b=-3$

$f(x) = x^2-3x+3$

$f(1) = 1-3+3 = 1$

Answer 1

Suppose $f$ has degree $n$, then we can write $$f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0$$ We are given that $f''(x)\to 2$ as $x\to\infty$, so what could $n$ possibly be?

Suppose $n>2$, then $\deg(f''(x))=n-2>0$, so $f''(x)\to\pm\infty$ as $x\to\infty$.

Suppose $n<2$, then $f''(x)\equiv 0$ , so clearly $f''(x)\to 0\neq 2$ as $x\to\infty$.

This should get you pretty close to a solution.

Answer 2

Dave showed that the degree of the polynomial is $2$. So we have $f(x)=a_2x^2+a_1x+a_0$.

We know that $f''(x)=2a_2=2\implies a_2=1\implies f(x)=x^2+a_1x+a_0$

We also know that there is a minima at $x=3\implies f'(3)=0\implies f'(3)=2\overbrace x^3+a_1=6+a_1=0\implies a_1=-6$ hence we have $f(x)=x^2-6x+a_0$.

We also know that $f(0)=3$ thus $f(0)={\overbrace x^0}^2-6\overbrace x^0+a_0=a_0=3$

And the final answer is $f(x)=x^2-6x+3$, with this we get $f(1)=1^2-6\times 1+3=\boxed{-2}$

Answer 3

Minima is attained at 3 , which means $\ f'(x) = 0$ at $\ x = 3$ . Use this to get correct result.