# continuous functions on intervals 2

Printable View

• Oct 29th 2008, 12:45 PM
rmpatel5
continuous functions on intervals 2
Let I:=[a,b] and let f:http://qaboard.cramster.com/Answer-B...4200002172.gif be a continuous function on I such that for each x in I there exists y in I such that http://qaboard.cramster.com/Answer-B...3575003345.gif. Prove that there exists a point c in I such that f(c)=0
• Oct 29th 2008, 12:56 PM
terr13
I think you can say something along the lines of take some f(x) in R, and w/ $d(0, f(x)) \leq \epsilon$. We can find a $y_1$ such that $d(0,f(y_1))\leq \epsilon /2$. Suppose there was minimal value c, such that $f(c) \geq 0.$ But we know we can find another value that is even smaller than c, and by the Archimedean Property, it must be 0. This is not a very rigorous proof though, since I didn't really use continuity.
• Oct 29th 2008, 01:00 PM
rmpatel5
Quote:

Originally Posted by terr13
I think you can say something along the lines of take some f(x) in R, and w/ $d(0, f(x)) \leq \epsilon$. We can find a $y_1$ such that $d(0,f(y_1))\leq \epsilon /2$. Suppose there was minimal value c, such that $f(c) \geq 0.$ But we know we can find another value that is even smaller than c, and by the Archimedean Property, it must be 0. This is not a very rigorous proof though, since I didn't really use continuity.

yeh i think it has to be more rigorous and using continuity. Any idea how to do it that way??
• Oct 29th 2008, 01:16 PM
Plato
There is a theorem say that “A continuous function on a closed interval has a minimum and a maximum”.
So $\left( {\exists m \in I} \right)\left( {\forall z \in I} \right)\left[ {f(m) \leqslant f(z)} \right]$.
So suppose that $\left( {\forall z \in I} \right)\left[ {0 \ne f(z)} \right]$ then from the given $f(m) > 0$.
A contradiction follows at once from the minimum.
• Oct 30th 2008, 03:49 PM
rmpatel5
Quote:

Originally Posted by Plato
There is a theorem say that “A continuous function on a closed interval has a minimum and a maximum”.
So $\left( {\exists m \in I} \right)\left( {\forall z \in I} \right)\left[ {f(m) \leqslant f(z)} \right]$.
So suppose that $\left( {\forall z \in I} \right)\left[ {0 \ne f(z)} \right]$ then from the given $f(m) > 0$.
A contradiction follows at once from the minimum.

I have been trying to do this problem as well along with the other problem, yet i still cant do it. So frustrated (Headbang). Being trying to read the book but I just dont understand this material.
• Oct 30th 2008, 04:03 PM
Plato
You are simply over-thinking all of this.
If the absolute minimum is $f(m)$ there cannot be any $y$ such that $\left| {f(y)} \right| \leqslant \frac{1}{2}\left| {f(m)} \right|$
Otherwise, $f(m)$ would not be the absolute minimum!