proof: working with derivatives, possibly mean value theorem?

• May 3rd 2010, 06:47 PM
sfspitfire23
proof: 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 PM
Drexel28
Quote:

Originally Posted by sfspitfire23
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?

First you need to realize what the question is asking.

If $\Gamma_f$ (the graph) intersects $\Delta=\left\{(x,x):x\in\mathbb{R}\right\}$ then that means that $(x,f(x))=(x,x)\implies f(x)=x$. So, why can't we have that $f(x)=x,f(x')=x'$ for $x\ne x'$?
• May 3rd 2010, 07:00 PM
sfspitfire23
Isn't x' just going to be zero, though, seeing as x is any real number?
• May 3rd 2010, 07:01 PM
Drexel28
Quote:

Originally Posted by sfspitfire23
Isn't x' just going to be zero, though, seeing as x is any real number?

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 $x_1,x_2,\text{ }x_1. Then, what does the MVT say about the derivative of $f$ on $[x_1,x_2]$?
• May 3rd 2010, 07:33 PM
sfspitfire23
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 PM
Drexel28
Quote:

Originally Posted by sfspitfire23
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) = $\color{red}1$
which contradicts the assumption that f'(x)>1 for all x...?

Yes, except I'm sure you meant what I changed above
• May 4th 2010, 06:03 AM
In more general terms, given $f'(x)\neq 1$, you can have at most one fixed point. But if you also have $|f'(x)|\leq A<1$, then you know for sure that there exists a fixed points, and hence it is unique. To show this, take the sequence defined by $x_{n+1}=f(x_n)$ and show it is convergent to the unique fixed points.