Expert: Paul Klarreich - 11/17/2009

Question
Hi I am not sure how to start this problem.

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 c between x1 and x2 for which
f(c) = [k1f(x1) + k2f(x2)]/ [k1+k2]

Thank You

Answer
Questioner: Sam
Country: United States
Category: Advanced Math
Private: No
Subject: Prove there exists a continuous function
Question: Hi I am not sure how to start this problem.

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 c between x1 and x2 for which
f(c) = [k1f(x1) + k2f(x2)]/ [k1+k2]

Thank You
..............................................
Hi,Sam,

You have the INTERMEDIATE VALUE THEOREM, that says:

IF

f(x) is continuous on [x1,x2], <<< never mind a,b.

AND

K is any number between  f(x1) and f(x2)

THEN

we can find  c between x1 and x2 such that  f(c) = K.

Now all you need do is show that your

   k1f(x1) + k2f(x2)
K = -----------------
        k1+k2

is between  f(x1) and f(x2).

Actually this expression is what is called a 'weighted average', but never mind that.

One of  f(x1),f(x2) must be bigger. [If they are equal, you can take c = either x1 or x2, which is no fun.]

Assume f(x2) > f(x1).  [It works the same the other way around.]

Now we shall prove that this K is between f(x1) and f(x2), namely:


f(x1) <= K <= f(x2)

K - f(x1) should be positive, and it is:

k1f(x1) + k2f(x2)
----------------- - f(x1) =
    k1+k2

k1f(x1) + k2f(x2) - (k1 + k2)f(x1)
----------------------------------- =
    k1+k2

k1f(x1) + k2f(x2) - k1f(x1) - k2 f(x1)
--------------------------------------- =
    k1+k2

k2f(x2) - k2 f(x1)
------------------- =
    k1+k2

k2[f(x2) - f(x1)]
------------------- =
    k1+k2

 k2
-----[f(x2) - f(x1)]
k1+k2

But f(x2) > f(x1), so [f(x2) - f(x1)] > 0

So K - f(x1) IS positive.

Likewise, you will show  f(x2) - K  is also positive.

and that will do it for you.