Suppose that f is ais continuos function defined on [0,1] with f(0)=1 and f(1)=0. show that there is a value of x that in [0,1] such that f(x)=x. Thank You.
This is slightly simpler:
Let's say I have a function g:[0,1] \rightarrow \Re, which is continuous, with g(0)=1 and g(1)=-1 can you show that there is an x \in [0,1] so that g(x)=0?
#3
jmich79
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Im Still Not Following. Can You Explain It A Little Bit Better To Me. Thank You For Your Post By The Way.
If a function that's continuous is negative at one point, and positive at another point, does it necesarily cross the x-axis (i.e. is zero somewhere in between)?
That's what he's driving at, but puts it in terms that are more obviously applicable to the problem at hand
Define the function G(x)=f(x)-x. Now four questions. i) is G continuous? ii) What are G(0) and G(1)? iii) What does it mean if G(x)=0 in terms of f? iv) Might this have something to do with the NateTG's and Office_Shredder's hints?
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...
Hello,
This is the attachment, the steps to solution are pretty clear. I guess there is a mistake on the highlighted part that prompts this thread.
Ought to be ##3^{n+1} (n+2)-6## and not ##3^n(n+2)-6##. Unless i missed something, on another note, i find the first method (induction) better than second one (method of differences).