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This originally was a quadratic question with $f(x)=ax^2 + bx + c$

If

$1 ≤ a + b + c ≤ 2$ {f(1)}

$2 ≤ 4a + 2b + c ≤ 3$ {f(2)}

$3 ≤ 9a + 3b +c ≤ 4$ {f(3)}

Prove $1 ≤ 16a + 4b + c ≤ 8$ {f(4)}

I think I got the limit of why it is $1$, but as for $8$ I am not too sure how to obtain it. Is it because its a quadratic it does that?

Itoz Darbien
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  • What do you mean by "I got the limit of why it is 1"? – Calvin Lin Jul 02 '23 at 00:37
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    To clarify, for the first line, are you trying to say $1 \leq f(1) \leq 2$, or $ 1 \leq a+b+c \leq 2f(1) $? (The latter doesn't seem true to me). – Calvin Lin Jul 02 '23 at 00:38
  • It must be $1\leq f(1) \leq 2$. – Itoz Darbien Jul 02 '23 at 00:39
  • Why did you delete your other identical copy of that question? https://math.stackexchange.com/questions/4729082/quadratics-with-range-of-function ? What is the point of posting more than once? I was also not aware of the existing answer here - had I been, I wouldn't have tried to reply to you because the answer here is quite good, to say the least! –  Jul 02 '23 at 08:51
  • sorry, i didnt understand this answer and you wrote the other one was a duplicate, so I deleted it –  Jul 02 '23 at 09:13

1 Answers1

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Hint: Prove that $f(4) - 3f(3) + 3f(2) - f(1) = 0 $ for a quadratic equation.

Hence, the desired inequality $1 \leq f(4) \leq 8$ follows.
When does / Can equality hold?

Calvin Lin
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  • what does that prove? –  Jul 02 '23 at 08:18
  • Well,@elementalmaths, if $f(4)=3f(3)-3f(2)+f(1)$ and $f(3)\le 4$ (so $3f(3)\le 12$), $f(2)\ge 2$ (so $-3f(2)\le -6$) and $f(1)\le 2$ then... –  Jul 02 '23 at 08:53