# Continuous function has only one local maximum, prove that it is its global maximum watch

1. Hi there,
If a continuous function with continuous derivatives has just one local maximum, how would we show that this local maximum is its global maximum?

Thanks!
2. (Original post by LaurenLovesMaths)
Hi there,
If a continuous function with continuous derivatives has just one local maximum, how would we show that this local maximum is its global maximum?

Thanks!
Is the question even true? What about f(x) = x^3-x?
3. (Original post by LaurenLovesMaths)
Hi there,
If a continuous function with continuous derivatives has just one local maximum, how would we show that this local maximum is its global maximum?

Thanks!
Have you tried using the definition of global maximum?
4. (Original post by RichE)
Is the question even true? What about f(x) = x^3-x?
Perhaps what's missing from the question is that there's only one local extremum and that the extremum is a local maximum.
5. (Original post by MR1999)
Perhaps what's missing from the question is that there's only one local extremum and that the extremum is a local maximum.
Yes sorry that is what I meant! The only critical point is the one maximum
6. (Original post by LaurenLovesMaths)
Yes sorry that is what I meant! The only critical point is the one maximum
Are you given any other information about the function?
7. (Original post by MR1999)
Are you given any other information about the function?
Its first and second derivatives are continuous (so f itself is continuous)
8. (Original post by LaurenLovesMaths)
Its first and second derivatives are continuous (so f itself is continuous)
Okay I think I got it. You have to prove that, if f(c) is the only extremum, then f is monotonic increasing on (-∞, c) and monotonic decreasing in (c, ∞) using the fact that f' is continuous and that f'(c)=0

Spoiler:
Show

This is my attempt:

We know that f has a one local extremum which is a maximum. This means that in some interval [a,b], there exists a c such that f(c)>f(x) ∀ x ∈ [a,b]. Also, it means that f is monotonic decreasing for x>c and monotonic decreasing for x>c, in the same interval. And given what we know about f', we also deduce that f'(x)>0 for x<c and f'(x)<0 for x>c and f'(x)=0 at x=c, again on the interval [a,b].

We can now extend these facts to Dom(ƒ) and claim that f(c) is the only extremum.

We prove this via contradiction. Suppose the above facts weren't true for Dom(f). Then we could say there exist some -∞<x<a such that ƒ'(x)<0. But there also exist a≤x<c such that f'(x)>0. So by the IVT, f'(x)=0 for some x∈(-∞,a]. This contradicts the fact that f only has one extremum. Applying the same argument for x>c completes the proof. □

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