Continuous not bounded above function

[mikael27]

Homework Statement

Let f : R -> R be a continuous function such that f(0) = 0. If S := {f(x) | x in R} is not
bounded above, prove that [0, infinity) ⊆ S (that is, S contains all non-negative real numbers).

Then find an appropriate value for a in the Intermediate Value theorem.

Homework Equations

The Attempt at a Solution

If y > 0, then since S is not bounded above, there exists b in R such that f(b) > y.

Then because y>0 , f(b) >0 and [0, infinity) ⊆ S ?

Applying the intermediate value theorem to this continuous function, it follows that there exists a real number a such that f(a) = y. ?

 

[xaos]

you should want to apply the IVT to the closed intervals contained in S of the continuous function.

 

[mikael27]

There is no given function. I think its more general question

 

[xaos]

perhaps i am misunderstanding a bit. first show that S is an interval (by continuity and IVT), then show by contradiction this interval has no upper bound.

 

[LCKurtz]

Homework Statement

Let f : R -> R be a continuous function such that f(0) = 0. If S := {f(x) | x in R} is not
bounded above, prove that [0, infinity) ⊆ S (that is, S contains all non-negative real numbers).

Then find an appropriate value for a in the Intermediate Value theorem.

Homework Equations

The Attempt at a Solution

If y > 0, then since S is not bounded above, there exists b in R such that f(b) > y.

That's true.

Then because y>0 , f(b) >0 and [0, infinity) ⊆ S ?

No, those two facts aren't enough, or even particularly relevant.

Applying the intermediate value theorem to this continuous function, it follows that there exists a real number a such that f(a) = y. ?

That last statement is true but you haven't given a convincing argument. Explain how your correct statement in red above combines with the IV theorem to get your last statement.

 

[mikael27]

how I am going to use the IVT with this statement?

 

[LCKurtz]

how I am going to use the IVT with this statement?

You are trying to show $y\in S$. You have b > 0 with f(b) > y and f(0)=0. So...

 

[Deveno]

f is continuous on the closed interval [0,b].

f(0) = 0

f(b) > y.

now ask yourself: can f get from f(0) to f(b) without crossing the horizontal line at y?

try drawing a picture, and re-read the statement of the IVT very carefully...

 

[mikael27]

so by IVT f(0)<=y<=f(b). then how we have to find the value for a using IVT

 

[Deveno]

what, exactly, does the IVT say? isn't it an existence theorem of some sort? what does it say exists?

(you are looking for a phrase such as: there is a c in [a,b] such that...)