# Intermediate Value Theorem problem

1. Sep 22, 2006

### mohlam12

I have a problem that I am stuck with, if anyone please can give me any hint, that'd be great.

f is a continuous function at the interval [a,b], and f(a) not euqal to f(b).
q and p are two positive integers, and they do not equal zero.

f : [a,b] ---> R

show that :

there is a number c that belongs to the interval [a,b] such as:
pf(a) + qf(b) = (p+q)f(c)
----
I only know that you should use the Intermediate Value Theorem, I only learned that this morning and I have no clue how to demonstrate this problem. Thanks for your help...

2. Sep 22, 2006

### StatusX

What does the intermediate value theorem say?

3. Sep 22, 2006

### mohlam12

" If f is continuous on a closed interval [a,b], and c is any number between f(a) and f(b) inclusive, then there is at least one number x in the closed interval such that f(x)=c. "

4. Sep 22, 2006

### istevenson

make a little change and it will be much easier to prove it.
step1
$$\frac{p}{p+q}f(a)+\frac{q}{p+q}=f(c)$$

step2
let $$\frac{p}{p+q}=\lambda$$
because p and q are all positive, so $$\lambda$$ is some value at the interval (0,1)
then we get $$\lambda f(a)+(1-\lambda)f(b)=f(c)$$

now you can use the Intermediate Value Theorem to prove it.

5. Sep 22, 2006

### mohlam12

I have to show an equality, I think that means I have to go from the first side, and reach the second part... and I still dont know how to prove your last line with the Intermediate Value Theorem

6. Sep 22, 2006

### istevenson

supposing that the f value is at the interval [M,N]
for $$\lambda$$ and $$1-\lambda$$ are positive,f(a) not eqal to f(b)
thus
$$\lambda f(a)+(1-\lambda)f(b) >\lambda M +(1-\lambda)M = M$$

$$\lambda f(a)+(1-\lambda)f(b) <\lambda N +(1-\lambda)N =N$$
so the value of $$\lambda f(a)+(1-\lambda)f(b)$$ is also at [M,N] according to the Intermediate Value Theorem,there exists some value at [a,b] that

$$f(c)=\lambda f(a)+(1-\lambda)f(b)$$

7. Sep 22, 2006

### mohlam12

okay
I came out to this problem now;
According to the IVT, knowing that f(a) not eqal to f(b), we have:
f(a) < f(c) < f(b)
so, f(a) is not equal to f(c), and, f(b) is not equal to f(c).
pf(a) not equal to pf(c), and qf(b) is not equal to qf(c).
if we add them together, we'll have:
pf(a) + qf(b) is not equal to (p+q)f(c)
and that doesn't not match with what I have to demonstrate.. ?!?!

8. Sep 22, 2006

### StatusX

But f(a)<(pf(a)+qf(b))/(p+q)<f(b). The IVT says any number satisfying this property has some c with f(c) equal to the number.