p(x) is a real polynomial of degree less than 3 and satisfies |p(x)| ≤ 1 for x ∈ [-1, 1]. Show that |p'(x)| ≤ 4 for x ∈ [-1, 1].
Solution
Straightforward.
Let p(x) = ax2 + bx + c, so p'(x) = 2ax + b. It is evidently sufficient to show that |2a + b| and |2a - b| ≤ 4. p(0) = c, p(1) = a + b + c, p(-1) = a - b + c, so 2a + b = 3/2 p(1) + 1/2 p(-1) + 2 p(0). But |p(1)|, |p(-1)|, |p(0)| ≤ 1, so |2a + b| ≤ 4. Similarly, 2a - b = 1/2 p(1) + 3/2 p(-1) + 2 p(0).
© John Scholes
jscholes@kalva.demon.co.uk
3 Nov 1999