Find max(abs(a)+abs(b)+abs(c)), and a possible f(x)

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Discussion Overview

The discussion revolves around finding the maximum value of the sum of the absolute values of the coefficients \(a\), \(b\), and \(c\) in the quadratic function \(f(x) = ax^2 + bx + c\), under the constraint that \(|f(x)| \leq 1\) for all \(x\) in the interval \([0, 1]\). Participants explore mathematical reasoning and inequalities related to this problem.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Exploratory

Main Points Raised

  • One participant states that by evaluating \(f(x)\) at specific points \(x = 0\), \(x = 1\), and \(x = \frac{1}{2}\), they derive expressions for \(f(0)\), \(f(1)\), and \(f\left(\frac{1}{2}\right)\), leading to inequalities involving \(|a|\), \(|b|\), and \(|c|\).
  • Another participant reiterates the same approach and concludes that \(|a| + |b| + |c| \leq 17\), based on the derived inequalities.
  • There is a mention of a potential typo in the previous posts, indicating a need for clarification or correction of the mathematical expressions presented.

Areas of Agreement / Disagreement

Participants appear to agree on the method of evaluating \(f(x)\) at specific points and the resulting inequalities, but the presence of a typo suggests that there may be some unresolved issues regarding the accuracy of the calculations or expressions used.

Contextual Notes

The discussion includes derived inequalities that depend on the assumptions made about the function \(f(x)\) and the specific values of \(x\) chosen for evaluation. The potential typo indicates that there may be limitations or errors in the mathematical steps presented.

Albert1
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$given\,\,f(x)=ax^2+bx+c$
$if \,\,for\,\,all\,\,x \in [0,1]\,\, we\,\, have\left | f(x) \right |\leq1$
$find \,\, max(\left | a \right |+\left | b \right |+\left | c\right |)$
$and \,\, a\,\, possible \,\, f(x)$
 
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Given $f(x) = ax^2+bx+c$ and $0 \leq x\leq 1$ and $|f(x)|\leq 1$Put $x=0$ in $f(x)\;,$ We get $f(0) = c$ and put $x=1\;,$ We get $f(1) = a+b+c$and put $\displaystyle x=\frac{1}{2}\;,$ We get $\displaystyle f\left(\frac{1}{2}\right) = \frac{a}{4}+\frac{b}{2}+c\Rightarrow 4f(1) = a+2b+4c$So Using $\triangle$ Inequality::$\displaystyle |a| = \mid 2f(1)-4f\left(\frac{1}{2}\right)+2f(0)\mid \leq 2|f(1)|+4\mid f\left(\frac{1}{2}\right)\mid+2|f(0)|\leq 8$$\displaystyle |b| = \mid-f(1)-4f\left(\frac{1}{2}\right)-3f(0)\mid \leq |f(1)|+4\mid f\left(\frac{1}{2}\right)\mid+3|f(0)|\leq 8$

$|c|\leq 1$So we get $|a|+|b|+|c| \leq 8+8+1 \leq 17$
 
jacks said:
Given $f(x) = ax^2+bx+c$ and $0 \leq x\leq 1$ and $|f(x)|\leq 1$Put $x=0$ in $f(x)\;,$ We get $f(0) = c$ and put $x=1\;,$ We get $f(1) = a+b+c$and put $\displaystyle x=\frac{1}{2}\;,$ We get $\displaystyle f\left(\frac{1}{2}\right) = \frac{a}{4}+\frac{b}{2}+c\Rightarrow 4f(1) = a+2b+4c$So Using $\triangle$ Inequality::$\displaystyle |a| = \mid 2f(1)-4f\left(\frac{1}{2}\right)+2f(0)\mid \leq 2|f(1)|+4\mid f\left(\frac{1}{2}\right)\mid+2|f(0)|\leq 8$$\displaystyle |b| = \mid-f(1)-4f\left(\frac{1}{2}\right)-3f(0)\mid \leq |f(1)|+4\mid f\left(\frac{1}{2}\right)\mid+3|f(0)|\leq 8$

$|c|\leq 1$So we get $|a|+|b|+|c| \leq 8+8+1 \leq 17$
one example of $f(x)$ : $a=8,b=-8,c=-1,\,\, we \,\, get \,\, f(x)=8x^2-8x-1$
$which \,\,max(|a|+|b|+|c|) =8+8+1 =17\,\,meets$
also for all $0 \leq x\leq 1 ,$ $|f(x)|\leq 1$
 
Last edited:
Albert said:
one example of $f(x)$ : $a=8,b=-8,c=-1,\,\, we \,\, get \,\, f(x)=8x^2-8x-1---(*)$
$which \,\,max(|a|+|b|+|c|) =8+8+1 =17\,\,meets$
also for all $0 \leq x\leq 1 ,$ $|f(x)|\leq 1$
a typo in $(*):$
$a=8,b=-8,c=1$
$f(x)=8x^2-8x+1$
$which \,\,max(|a|+|b|+|c|) =8+8+1 =17\,\,meets$
also for all $0 \leq x\leq 1 ,$ $|f(x)|\leq 1$
 

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