max min (1 Viewer)

[paper]

Given lim (x -> inf) f(x) = lim (x -> -inf) f(x) = 0, and f(x) continuous over R,
Show that (editif there is a number z such that f(z) > 0 then f attains a maximum value on R.

I know why intuitively but i dunno how 2 express it mathematically. some help would b greatly appreciated. thanx

 

[turtle_2468]

Is this an assignment question? If not I can put up the soln tomorrow..

 

[paper]

no, just 1 q out of many homework exercises.

edit: btw will there be assessable h/w assignments consisting of maths qs in 2nd yr maths?

 

[CM_Tutor]

Originally posted by paper
Given lim (x -> inf) f(x) = lim (x -> -inf) f(x) = 0, and f(x) continuous over R,
Show that there is a number z such that f(z) > 0 then f attains a maximum value on R.

There is a problem with this question as it stands, as it isn't true. Consider the function f(x) = -1 / (1 + x²). This has no maximum, and is never positive. It is continuous and differentiable for all x in R, and as x --> +/- inf,
f(x) --> 0^-. Certainly, it is bounded above by 0, but it never attains a maximum.

 

[wogboy]

There is a problem with this question as it stands, as it isn't true. Consider the function f(x) = -1 / (1 + x2). This has no maximum, and is never positive. It is continuous and differentiable for all x in R, and as x --> +/- inf,
f(x) --> 0-. Certainly, it is bounded above by 0, but it never attains a maximum.

paper must have forgotten to put the word "if" between the words "that" & "there"

Show that there is a number z such that f(z) > 0 then f attains a maximum value on R.

 

[CM_Tutor]

You're right, it is true of ot says 'show that if there is a number z such that f(z) > 0, then f attains a maximum value on R. Note that this maximum value need not be a stationary point.

 

[Affinity]

Let me save turtle some work:
Cut the infinite interval down to a finite interval:
f(z) > 0

let E= f(z), there must exist c,d such that whenever x < c or when x > d, f(x) < E by the definition of the limits.

consider the interval [c,d], there must be a maximum in the interval by the max min theorem, let it be M.

M> {f(x):c<=x<=d}
M > f(z) because z is in [c,d]
M > {f(x):x&lt;c OR &gt;d} because f(z) = E is more than those values.

it follows that M is a maximum over R.

 

[paper]

oops 4got the if...

thanx 4 that affinity, just one q though... doesnt the third case assume that f(x) > E occur continuously on only one interval [c,d]? Would it b correct 2 say that there a may be a multitude of individual intervals that satisfies the criterium that defined [c,d] and each having its own maximum M_n and the there must exist a maximum value over R that is equal 2 max {M₁, M₂, ... , M_n} ?

 

[turtle_2468]

Paper: No.
There are a few intervals which satisfy criterium defining [c,d] etc (eg just by making d larger)... but suppose you have two of these intervals (say [3,5] and [2,6] - too late to algebra this rigorously but I think this is just to satisfy you anyway.. )
Then the max M_2 of [2,6] will be larger than or equal to E. Suppose this is in fact larger than M_1 the max in [3,5]. Then a value larger than or equal to E occurs in [2,3) or (5,6]. But this results in a contradiction (because of defn of continuity applied to [3,5]).
QED. Of course, subcases where intervals don't have one inside another etc. but rough idea is same. If occurs must occur in first interval meaning same max.

 

[paper]

ahhhh.... i see... i understand now. thanks heaps

 

[giraffe]

Believe it or not, I have a friend named Max Min!!

Originally posted by paper
ahhhh.... i see... i understand now. thanks heaps