Try this beautiful Functional Equation Problem from SMO, Singapore Mathematics Olympiad, 2013.
Let M be a positive integer .It is known that whenever \(|ax^2 + bx +c|\leq 1\) for all
\(|x|\leq 1\) then \(|2ax + b |\leq M \) for all \(|x|\leq 1\). Find the smallest possible value of M.
Functional Equation
Function
Answer: 4
Singapore Mathematics Olympiad
Challenges and Thrills - Pre - College Mathematics
We cant this sum by assuming a,b,c as fixed quantity.
Let \( f(x) = ax^2 + bx + c \).
Then \( f(-1) = a - b + c \) ; \( f(0) = c \) ; \( f(1) = a + b + c\) ;
Try to do the rest of the sum ................................
Suppose \( |f(x)|\leq 1\) for all \(|x|\leq 1 \) . Then
\( |2ax + b| = | (x - \frac {1}{2} ) f(-1) - 2 f(0) x + (x+\frac {1}{2} f(1) |\)
\(\leq |x - \frac {1}{2}| + 2 |x| + |x + \frac {1}{2}|\)
\(\leq |x - \frac {1}{2} | + |x+\frac {1}{2}| + 2 \)
\(\leq 4 \)
Now I guess you have already got the answer but if not .............
From the last step we can conclude ,
\(|2 x^2 - 1|\leq 1 \) whenever \(|x|\leq 4\) and \(|2x| = 4 \)
is achieved at \(x = \pm 1\).
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