# MVT question + fixed points

1. May 2, 2010

### nugget

1. The problem statement, all variables and given/known data

Consider the function g: [0,∞) -> R defined by G(x)=x+e-2x

A. Use the mean value theorem to prove that |g(x2)-g(x1)|<|x2-x1| for all x1,x2 E [0,∞) with x1≠x2.

B. Find all fixed points of g on [0,∞).

2. Relevant equations

MVT: f'(c) = f(b)-f(a)/b-a.

3. The attempt at a solution

It seems we are being asked to prove in part A. that |g(x2)-g(x1)|/|x2-x1|<1.

This should be proved if we show that g'(x)<1 for x E [0,∞).

g'(x) = 1-2e-2x. By inspection g'(x) is clearly less than 1 for x values between 0 and ∞.

Is this sufficient an explanation? M.V.T. states that the gradient of a function must, at at least one point (c,f(c)) between the points (a,f(a)) and (b,f(b)), equal the slope over the whole region [a,b] (as the equation shows). My method is essentially the same but replaces the 'rise over run' bit of the MVT simply with f'(x).

Am I on the right track, is there a better way to explain this, perhaps more directly involving MVT...? Thanks

have not yet tackled question B but any hints would be appreciated