SOLUTION TO PROBLEM 9

For a real number $a\in {\mathbb R}$ we define a function $g_a:{\mathbb R} \longrightarrow {\mathbb R}$ by

$$g_a(x)=f'(x+a)\sin(x)-f(x+a)\cos(x).$$

Then, for every $x\in [0,\pi]$, we have

$$g'_a(x)=\sin(x)\cdot \big(f(x+a)+f''(x+a)\big)\geq 0,$$

and therefore g_a is a non-decreasing function. Hence, for every $a\in {\mathbb R}$,

$$0\leq g_a(\pi)-g_a(0)=f(\pi+a)+f(a)=2\cdot f(a),$$

as required.
 
   
 

Questions and/or comments should be directed to Judy Downey or Griff Elder
 

    Back to the Problem of the Week Page
 
 

Last modified:   Tue Mar 13 09:49:05 CST 2001