Surjective Proof Homework: Show f is Surjective on (c,d)

In summary: Basically, you want to show that f is continuous, and therefore takes on all values between its minimum and maximum on the interval (a,b). This means that f is surjective on some open interval (c,d), where c and d are the minimum and maximum values of f on (a,b).
  • #1
analysis001
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Homework Statement


Suppose f: (a,b)→R where (a,b)[itex]\subset[/itex]R is an open interval and f is a differentiable function. Assume that f'(x)≠0 for all x[itex]\in[/itex](a,b). Show that there is an open interval (c,d)[itex]\subset[/itex]R such that f[(a,b)]=(c,d), i.e. f is surjective on (c,d).


Homework Equations


f is surjective if for all y[itex]\in[/itex]R there exists an x[itex]\in[/itex]X such that f(x)=y.


The Attempt at a Solution


I think I'm supposed to use ε and δ for this proof but I'm not sure where to start. Any clues would be great! Thanks.
 
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  • #2
analysis001 said:

Homework Statement


Suppose f: (a,b)→R where (a,b)[itex]\subset[/itex]R is an open interval and f is a differentiable function. Assume that f'(x)≠0 for all x[itex]\in[/itex](a,b). Show that there is an open interval (c,d)[itex]\subset[/itex]R such that f[(a,b)]=(c,d), i.e. f is surjective on (c,d).

Homework Equations


f is surjective if for all y[itex]\in[/itex]R there exists an x[itex]\in[/itex]X such that f(x)=y.

The Attempt at a Solution


I think I'm supposed to use ε and δ for this proof but I'm not sure where to start. Any clues would be great! Thanks.

I don't think you need ε and δ. Start by thinking about continuous functions (like f, since it's differentiable) and the Intermediate Value Theorem. The Mean Value theorem will come in handy too.
 
Last edited:

1. What does it mean for a function to be surjective?

A function is considered surjective if every element in the range of the function has at least one corresponding element in the domain. In other words, every output value is mapped to at least one input value.

2. How can I prove that a function is surjective?

To prove that a function is surjective, you need to show that for every element in the range, there is at least one element in the domain that maps to it. This can be done by using an arbitrary element in the range and finding its corresponding element in the domain.

3. What is the difference between a surjective function and an injective function?

A surjective function maps every element in the range to at least one element in the domain, while an injective function maps each element in the range to a unique element in the domain. In other words, a surjective function is "onto" while an injective function is "one-to-one".

4. Can a function be both surjective and injective?

Yes, a function can be both surjective and injective. This type of function is called a bijective function and it means that every element in the range is mapped to a unique element in the domain, and every element in the domain has a corresponding element in the range.

5. How does proving that a function is surjective relate to its inverse function?

If a function is surjective, then its inverse function exists. This means that the inverse function can map every element in the range of the original function back to the corresponding element in the domain. Proving that a function is surjective is an important step in showing that its inverse function exists.

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