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

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Homework Statement


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


Homework Equations


f is surjective if for all y\inR there exists an x\inX 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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analysis001 said:

Homework Statement


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

Homework Equations


f is surjective if for all y\inR there exists an x\inX 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.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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