Expert: Alon Mandes - 11/12/2009

Question
Suppose that the function f is continuous on [a,b], x1 and x2 are in [a,b] and k1 and k2 are positive real numbers.  Prove that there exists a c between x1 and x2 for which:

f(c)= (k1f(x1)+k2f(x2)/k1+k2)

when I put x1 or k1 it means x sub 1 or k sub 1.

I was thinking that I would use mean value theorem.  Would I start my proof out like this

x1≤(k1f(x1)+k2f(x2)/k1+k2)≤x2  assuming that x1 is less than x2

Any help wouuld be great!

Answer
No, No M.V.T is needed. %26 besides M.V.T concerns the derivative , so it's irrelevant.
What we might do here is to start with simple proof :

Suppose a strait line on x-axis with ends a %26 b . Where a<b . For every real positive k1,k2 :
a ≤ (k1a+k2b)/(k1+k2) ≤ b .
Proof :
Left side : a ≤ (k1a+k2b)/(k1+k2) --> a(k1+k2) ≤ k1a+k2b --> ak1+ak2 ≤ ak1+bk2 --> a ≤ b . True
Right side: (k1a+k2b)/(k1+k2) ≤ b --> k1a+k2b ≤ b(k1+k2) --> k1a+k2b ≤ bk1+bk2 --> a ≤ b . True
End. Q.E.D

Now, we will apply this result to y-axis, where the points on the y-axis is value of functions .

1st case : Function f is increasing in the interval [a,b]
---------------------------------------------------------
Suppose that x1<x2 then f(x1)<f(x2) .
Lets set α=f(x1) %26 β=f(x2) . Thus , According to our previous result :
Every real k1 %26 k2 determine γ , such as : α ≤ (k1α+k2β)/(k1+k2) ≤ β Since the function f is
continues in the interval [a,b] . %26 because f is continues this means that γ=f(c) where
x1 ≤ c ≤ x2 . Therefore : α ≤ γ ≤ β , which leads to f(x1) ≤ γ ≤ f(x2) %26 hence,
γ=f(c)=(k1f(x1)+k2f(x2)/k1+k2). Q.E.D


2nd case : Function f is decreasing in the interval [a,b]
---------------------------------------------------------
Suppose that x1<x2 then f(x2)<f(x1) .
Lets set α=f(x1) %26 β=f(x2) . Thus , According to our previous result :
Every real k1 %26 k2 determine γ , such as : β ≤ (k1α+k2β)/(k1+k2) ≤ α Since the function f is
continues in the interval [a,b] . %26 because f is continues this means that γ=f(c) where
x1 ≤ c ≤ x2 . Therefore : β ≤ γ ≤ α , which leads to f(x2) ≤ γ ≤ f(x1) %26 hence,
γ=f(c)=(k1f(x1)+k2f(x2)/k1+k2). Q.E.D


Alon.