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 ).
It may be a chanllenge problem, please help me out.
Thank you more than I can say!
What is the middle value theorem? That may give some insight as to how to solve this. The is what makes this bothersome of course, for if that stipulation wasn't in place we know there exists some s.t. . Which of course taking makes this trivial. I'll give it more thought.
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?
By the mean value theorem, there is a point c with 0<c<1 such that . If there is more than one such point, say c<d and then clearly we can take and . So we may assume that there is a unique point c with .
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 . (A similar proof would deal with the case .)
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 so that . Then the line will have those properties. In fact, if we define then you can verify that and . 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 with with . Therefore .
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 . Then the line 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 with such that .
Then .
[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).]
I had overlooked that possibility. But in fact the proof becomes easier in that situation, because 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.