Expand the last inequality. Add f(c) to all three parts. See what you get.
Hi, I'm stuck on the following question. Any help will be fantastic, thanks.
is a continuous function. Prove that if for some , , then there exists a such that ,
I tried a contradiction proof but didn't really get anywhere. Please help! Thank you.
Following your help, I get...
So and ...each of which is greater than 0 since f(c)>0
Is this what you meant? I don't see where the third inequality comes from so I don't think I've followed your advice properly.
Also, if a function is continuous at c, by definition, , there exists a s.t. such that , .
As far as I can see, you've let which is fixed. So I don't see how this covers the 'for all epsilon' bit. I'm probably being very dull here though.
EDIT: Thinking about this again, I think you might've been suggesting a proof by contradiction - which I totally missed. If you let f(x) < 0, then we have a contradiction. Is that what you wanted me to do?
First of all, there is absolutely no reason why we should for “all epsilon”.
The problem says that f is positive is some neighborhood of c.
Use the definition of continuity with .
We know that
We also know that which is a neighborhood of c. For any x there we have
So the function is positive in that neighborhood of c.