Extreme and Intermediate value theorem

  • Thread starter Thread starter mikael27
  • Start date Start date
  • Tags Tags
    Theorem Value
mikael27
Messages
59
Reaction score
0

Homework Statement



Let f : [a; b] ! R be an arbitrary continuous function. Let S = {f(x)| a<= x<=b}. Show
that if S contains more than one element, then S is an interval of the form [c, d].

Hint: First apply the Extreme Value theorem, then the Intermediate Value theorem.

Homework Equations





The Attempt at a Solution



dont have any clue

 
Physics news on Phys.org
If we suppose S contains more than one points then are a \le c_1 &lt; c_2 \le b such that f(c_1) \neq f(c_2). Now the EVT can be applied to say something about the relationship of these. Once that's established the MVT will show that it must be an interval.

This actually says something quite important about continuous mappings over real numbers.
 
In response to your PM, the EVT can be applied to say that, in addition (without loss of generaltiy) c_1 and c_2 are the minimum and maximum on this interval, respectively. We can say this because continuous function must attain their maximum and minimum.

Now the IVT can be applied to show that f also attains all values between f(c_1) and f(c_2). You can do the same for the intervals [a,c_1] and [c_2, b], then you have f(c_2) = d &gt; c = f(c_1) and so f([a,b]) = [c,d].


It stil needs some details, but that's the gist of it.
 
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...

Similar threads

Back
Top