Expert: Paul Klarreich - 2/16/2007

Question
p(x) is a POLYNOMIAL of degree less than 3.
|p(x)|<=1  for all x in [-1,1].

show that:
|p'(x)|<=4 for all x in[-1,1].

Answer
Questioner:   Siddharth
Category:  Calculus
 
Question:  p(x) is a POLYNOMIAL of degree less than 3.
|p(x)|<=1  for all x in [-1,1].

show that:
|p'(x)|<=4 for all x in[-1,1].
 
....................................
Hi, Siddarth,

You have a graph that is totally contained in the square box with vertices at (+-1,+-1).

Suppose  p(x) has degree 1. (Linear)  Then it's easy to prove that the maximum slope is equal to 1.  So assume that p(x) is quadratic.

p(x) = ax^2 + bx + c,  a /= 0

p'(x) = 2ax + b

To achieve maximum slope, you want the graph to go as high as possible, so assume that it rises to (1,1) at the right, and you want it to go as low as possible, so assume that its vertex lies on the line  y = -1.  The maximum slope is achieved at the right, so we will compute the slope as a function of the location of the vertex, which depends on a and b.

The x-coordinate of the vertex is at -b/2a.

We will maximize  p'(1) = 2a + b.

Do this as a standard max-min problem.  You have a quantity to be optimized subject to constraints.

The problem could be written:

Maximize the slope of the parabola y = ax^2 + bx + c at the point (1,1), which is:

m = 2ax + b, at (1,1), which is  m = 2a + b

Constraints:

A. The graph must pass through (1,1).
B. The vertex must be on the line  y = -1.
C. The graph never goes above  y = 1.

Condition A means  a + b + c = 1.
Condition B: The x-coordinate of the vertex is at -b/2a.
The y-coordinate is  a(-b/2a)^2 + b(-b/2a) + c = -1

b^2/4a - b^2/2a + c = -1
      - b^2/4a + c = -1
Combine that with A:

      - b^2/4a + c = -1
         a + b + c =  1
--------------------------
a + b + b^2/4a = 2
Clear:
4a^2 + 4ab + b^2 = 8a
(2a + b)^2 = 8a

2a + b = 2 sqrt(2a)

So  m = 2 sqrt(2a)

dm/da = 2(1/2)(2a)^-1/2(2) = 2/sqrt(2a)

How do we use that?  Since we always have  a > 0 on the assumption that the graph is rising, this is never zero, so there is no stationary point.  The maximum will have to come from an endpoint of the appropriate interval for values of a.

Since
(1) by condition C, the vertex cannot be located to the right of  x = 0, and  
(2) the 'steepness' of the graph increases as the vertex moves to the right, the maximum slope must be attained by the parabola that has its vertex at x = 0. So we can determine the exact equation.  

Since  -b/2a = 0, b = 0.  We have  y = ax^2 + c

Since f(0) = -1,  c = -1. We have  y = ax^2 - 1

Since f(1) = 1,  a(1)^2 - 1 = 1, and a = 2

So the equation is  y = 2x^2 - 1

m = 4x

And the maximum slope occurs at x=1:  |m| = |4(+-1)| = 4.