If $f(x)=ax^2+bx+c$ be such that $|f(0)|\leq 1$,$|f(1)\leq 1$ and $|f(-1)\leq 1$, then for $x\in [-1,1]$ then what may be maximum value of $|f(x)|$

Asked Aug 02 '22 at 14:56

Active Aug 02 '22 at 17:19

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If $f(x)=ax^2+bx+c$ be such that $|f(0)|\leq 1$,$|f(1)|\leq 1$ and $|f(-1)|\leq 1$, then for $|x|\leq 1$, $|f(x)|$ cannot have the value

(A) $\frac{1}{4}$

(B) 1

(D) $\frac{7}{4}$

My Attempt

I am trying to fix $x=-\frac{b}{2a}$ between $-1$ and $1$ so that all conditions may get satisfied but am not able to reach any conclusion

edited Aug 02 '22 at 17:19

GaloisTH

asked Aug 02 '22 at 14:56

Maverick

Can't vertex lie outside $[-1,1]$? – aarbee Aug 02 '22 at 15:01
Why is there a close vote to it? – aarbee Aug 02 '22 at 15:02
2

I think this is same as the one here https://math.stackexchange.com/questions/774069/let-fx-ax2bxc-where-a-b-c-are-real-numbers-suppose-f-1-f0-f1-i?rq=1 – Maverick Aug 02 '22 at 15:04
Yes. My question is answered here – Maverick Aug 02 '22 at 18:13

0 Answers0