Real analysis, simpl(er) questions

[rayred]

Homework Statement

It is a 4 parter, but i got 3 and 4 done.

a) Find f ([0,3]) for the following function:
f(x)=1/3 x^3 − x + 1

b) Consider the following function :
f(x) = e^(−ax) (e raised to the power of '-a' times 'x') a, x ∈ [0,∞)
Find values of a for which f is a contraction .

The Attempt at a Solution

You would think that a is simple to me, but it is not. How does one go about solving a, because I have in my notes to take the max and min of [0, 3] then evaluate at those points, then do something with f prime? I am all confused because I must have screwed up my notes. Should not be too hard to answer

Now, I think I know what a contraction is, but I seem to be having problem
A contraction is something defined as such:

| f(x) - f(y) | <= n|x - y| for some 0 < n < 1... correct?
am i setting
f(x) = e^(-a_1x)
f(y) = e^(-a_2x)
so are we finding a value a that ensures 0 < n < 1 ? if so how?

 

[lippyka]

a) f([0,3])=1/3 (3^3)-(3)+1 = 8b) To find the values of 'a' for which f is a contraction, we need to set up the equation. We have |f(x)-f(y)|<=n|x-y|. We can substitute f(x) and f(y) with e^(-ax) and e^(-ay), respectively. This gives us |e^(-ax)-e^(-ay)|<=n|x-y|. After rearranging, this becomes e^(-ax)-e^(-ay)<=n|x-y|. Taking the natural log of both sides, we get -ax-ay<=ln(n)|x-y|. Simplifying, we get -(a)(x+y)<=ln(n)|x-y|. This can be further simplified to a=-ln(n)/(x+y). Thus, the value of 'a' that ensures 0 < n < 1 is ln(n)/(x+y).