Let f:R-->R be differentiable for all x. Suppose that f'(x)>1 for all x. Show that the graph of y=f(x) can intersect the graph of y=x no more than once.

I have no idea where to even begin... any ideas?

- May 3rd 2010, 06:47 PMsfspitfire23proof: working with derivatives, possibly mean value theorem?
Let f:R-->R be differentiable for all x. Suppose that f'(x)>1 for all x. Show that the graph of y=f(x) can intersect the graph of y=x no more than once.

I have no idea where to even begin... any ideas? - May 3rd 2010, 06:49 PMDrexel28
- May 3rd 2010, 07:00 PMsfspitfire23
Isn't x' just going to be zero, though, seeing as x is any real number?

- May 3rd 2010, 07:01 PMDrexel28
I have no idea what you mean. This is much, much simpler than you're making it. The question is asking why a function whose derivative which is strictly greater than one can't have two fixed points. To see this suppose that it did, call them . Then, what does the MVT say about the derivative of on ?

- May 3rd 2010, 07:33 PMsfspitfire23
Oh I see now. So if we suppose towards contradiction that f has two fixed points: x_1 and x_2 with x_1< x_2, then we would have f(x_1)= x_1 and f(x_2)= x_2. Then, by the Mean Value Theorem, we would have for some c in the interval (x_1, x_2):

f'(c) = f(x_2) - f(x_1) / x_2 - x_1 = (x_2 - x_1)/(x_2 - x_1) = 0

which contradicts the assumption that f'(x)>1 for all x...? - May 3rd 2010, 08:09 PMDrexel28
- May 4th 2010, 06:03 AMmohammadfawaz
In more general terms, given , you can have at most one fixed point. But if you also have , then you know for sure that there exists a fixed points, and hence it is unique. To show this, take the sequence defined by and show it is convergent to the unique fixed points.