Recent content by james007

Homework Statement A pen falls from the top of a roof. We know that it takes 1.5 seconds to fall the last 1/4 distance. Find the height of the roof from the ground. Homework Equations x=x_0 + v_0 * t + 1/2 gt^2 v=v_0 + gt The Attempt at a Solution Since the problem does not say...

Thank you mathwonk for a very helpful and informative thread. I should have seen this thread earlier so that I can ask you questions about algebra. I have no intention to become a mathematician because I think it's too hard for me, but I plan to obtain a master degree in pure/applied math. I...

You told me to consider the function g(x) = f(x) -x. So, if there exists x such that f(x) =x , then g(x) must be equal to 0 for some x. By contradiction, I assume that g(x) doesn't equal 0 for all x. Pick x= 0 , g(0) does not equal 0 by the assumption I just made. There are two cases: g(0) < 0...

Sorry, I meant decreasing. A typo. So we have [ g(x) - g(0) ] / x is approaching 0 this means there exists some y such that g'(y) = 0. A contradiction since above we assume that g'(x) < 0 for all x. Can you help me with the counterexample when M=1?

Yes, it must cross the line y=x. So from here I get the existence of x where f(x)= x ?

No, if g' is always negative, then g is increasing. What does this say about f(x)?

I don't quite get the term "global behavior" . Can I say that the graph of g assumes max and min?

Derivative of g implies that f '(x) - 1 < M , which I already stated above. But the inequality f '(x) < M < 1 implies that f '(x) - 1 < M -1 < 0 right?

If there were two such points x, y. Wlog, let x < y, then by MVT, there exists z in (x,y) such that f(y) -f(x) = f '(z) * (y-x) . Then for each z , there exists M < 1 such that f '(z) < M . So, f(y) - f(x) < M * (y-x) , i.e.( ( f(y) - f(x) ) / (y-x) ) < M ?

Thanks for your help. I just have no clue how to approach this problem. By assumption, we have for each x in R, there exists M < 1 such that f '(x) < M . I just don't know what to do with this given info. Can you elaborate a bit? Consider g(x) = f(x) -x , then g(x) is also differentiable. So...

Can someone help me with this? Let f: R into R be differentiable 1) If there is an M strictly less than 1 for each x in R, f'(x) strictly less than M, prove that there exists a unique point x such that f(x)=x. ( Note: x is a fixed point for f) 2) Give counter example to show 1) fails if M=1.