# Thread: continuous functions on intervals 2

1. ## continuous functions on intervals 2

Let I:=[a,b] and let f: be a continuous function on I such that for each x in I there exists y in I such that . Prove that there exists a point c in I such that f(c)=0

2. 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.

3. 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??

4. 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.

5. 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 . Being trying to read the book but I just dont understand this material.

6. 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!