# Thread: Help! The middle value theorem

1. ## Help! The middle value theorem

f(x) is continous in [0,1], differentiable in (0,1), and f(0)=0,f(1)=1.
prove that for any positive numbers a and b, there exist < in (0,1) such that
.

P.S I have posted this thread in the calculus subforum, but it is unsolved yet. I just want somebody help me out.

2. We have never met problem like this, (usually we are supposed to provethat there is a point in the open interval satisfies certain condition which contains only one parameter x ).
Thank you more than I can say!

3. What is the middle value theorem? That may give some insight as to how to solve this. The $\displaystyle x_1<x_2$ is what makes this bothersome of course, for if that stipulation wasn't in place we know there exists some $\displaystyle c\in(0,1)$ s.t. $\displaystyle f'(c)=\frac{1-0}{1-0}=0$. Which of course taking $\displaystyle x_1=x_2=c$ makes this trivial. I'll give it more thought.

4. Originally Posted by Drexel28
What is the middle value theorem? That may give some insight as to how to solve this. The $\displaystyle x_1<x_2$ is what makes this bothersome of course, for if that stipulation wasn't in place we know there exists some $\displaystyle c\in(0,1)$ s.t.
$\displaystyle f'(c)=\frac{1-0}{1-0}=0$. here 1, not 0.

Which of course taking $\displaystyle x_1=x_2=c$ makes this trivial. I'll give it more thought.
A type error!

5. Originally Posted by Shanks
A type error!
Ok? Sweet. That was an obvious typo. That does not in anyway answer the question.

6. I mean the intermediate value theorm.
we can give a little manipulation to the representation, that is equivalent to show that
for any positive number $\displaystyle \lambda$ in (0,1), there exist $\displaystyle x_1<x_2,\text{ such that }$$\displaystyle \frac{\lambda}{f'(x_1)}+\frac{1-\lambda}{f'(x_2)}=1$

7. Is this statement true or false?
"suppose f is continous in [a,b], differentiable in (a,b). If r and s in (a,b) satisfy f'(r)<f'(s), then for any point c in the open interval ( f'(r), f'(s) ), there exist a point t in (a,b) such that f'(t)=c."
I think it is true! and we may apply it to the prove. what is your opinion?

8. Originally Posted by Shanks
f(x) is continous in [0,1], differentiable in (0,1), and f(0)=0,f(1)=1.
prove that for any positive numbers a and b, there exist < in (0,1) such that
.
By the mean value theorem, there is a point c with 0<c<1 such that $\displaystyle f'(c)=1$. If there is more than one such point, say c<d and $\displaystyle f'(d)=1$ then clearly we can take $\displaystyle x_1=c$ and $\displaystyle x_2=d$. So we may assume that there is a unique point c with $\displaystyle f'(c)=1$.

It cannot happen that f(c)=c (because then the mean value theorem applied to the interval [0,c] would lead to another point where the derivative is 1. So suppose that $\displaystyle c<f(c)<1$. (A similar proof would deal with the case $\displaystyle 0<f(c)<c$.)

The idea now is to draw a line through the point (c,f(c)) with gradient less than 1 but large enough that its value at x=1 is larger than 1. This line must cross the graph of f(x) at some point between c and 1. To do that, choose $\displaystyle \delta>0$ so that $\displaystyle \delta<\frac{f(c)-c}{a(1-f(c))}$. Then the line $\displaystyle y-f(c) = \frac{x-c}{1+a\delta}$ will have those properties. In fact, if we define $\displaystyle g(x) = f(x) - f(c) - \frac{x-c}{1+a\delta}$ then you can verify that $\displaystyle g'(c)>0$ and $\displaystyle g(1)<0$. So g(x) must be positive when x is slightly larger than c, and g(x) is negative when x=1. By the intermediate value theorem there is a point q with c<q<1 such that g(q)=0. Then by Rolle's theorem there is a point $\displaystyle x_2$ with $\displaystyle c<x_2<q$ with $\displaystyle g'(x_2)=0$. Therefore $\displaystyle f'(x_2) = \frac1{1+a\delta}$.

Now we do something similar in the interval [0,c]. This time, we want a line through (c,f(c)) with gradient greater than 1 that crosses the y-axis above the origin. To do that, we choose $\displaystyle \delta<\frac{f(c)-c}b$. Then the line $\displaystyle y-f(c) = \frac{x-c}{1-b\delta}$ has the required properties, and it must cross the graph of f(x) at some point p between 0 and c. A similar argument to the previous paragraph shows that there is a point $\displaystyle x_1$ with $\displaystyle 0<x_1<c$ such that $\displaystyle f'(x_1) = \frac1{1-b\delta}$.

Then $\displaystyle \frac a{f'(x_1)} + \frac b{f'(x_2)} = a(1-b\delta) + b(1+a\delta) = a+b$.

[If we were told that f'(x) is continuous then the above argument could be made a lot shorter, by applying the intermediate value theorem to f'(x). But we are not told that, and so the only tool available is to use the mean value theorem on f(x).]

9. I think your prove have some flaw.

10. Originally Posted by Shanks
I think your prove have some flaw.
I had overlooked that possibility. But in fact the proof becomes easier in that situation, because $\displaystyle \delta$ can then take any positive value and the rest of the argument works as before.

I was not trying to provide a watertight analytic proof of this rather tricky result, but to give an indication of the geometric reason why it is true. If f(c)>1 then any line through (c,f(c)) with positive gradient less than 1 will cross the graph of f(x) between x=c and x=1.

11. Originally Posted by Opalg
If we were told that f'(x) is continuous then the above argument could be made a lot shorter, by applying the intermediate value theorem to f'(x). But we are not told that, and so the only tool available is to use the mean value theorem on f(x).
Actually you don't need that f'(x) is continuous. It is know that if $\displaystyle f'(a)<\lambda<f'(b)$ then there exist a $\displaystyle c\in(a,b)$ such that $\displaystyle f'(c)=\lambda$

12. Originally Posted by putnam120
Actually you don't need that f'(x) is continuous. It is know that if $\displaystyle f'(a)<\lambda<f'(b)$ then there exist a $\displaystyle c\in(a,b)$ such that $\displaystyle f'(c)=\lambda$
Thanks, that's something I should have known (probably did, at some time in the remote past, but my memory's not what it was). In fact, I think that the technique I used in the above comment would probably also prove that interpolation property for the derivative.

13. So...the statement is true. right?