# Thread: Two questions on functions

1. ## Two questions on functions

Okay, here is one:

Suppose that f:R-->R is continuous for all x and let alpha be in R. Show that if there is a b in R such that f(b)>alpha, then there is a delta>0 so that for all x such that |x-b|<delta, we have f(x)>alpha

So far, I know that you're going to have to make use of the definition of continuity. THat's about as far as I got though.

Another:

Suppose that f:R-->R is continuous for all x and let f(q)=0 for all rationals q. Prove that f(x)=0 for all x

2. Originally Posted by jmoney90
Suppose that f:R-->R is continuous for all x and let alpha be in R. Show that if there is a b in R such that f(b)>alpha, then there is a delta>0 so that for all x such that |x-b|<delta, we have f(x)>alpha
In the definition of continunity let $\displaystyle \varepsilon = \frac{{f(b) - \alpha }}{2} > 0$.

Now the way this forum works, you should show some effort on your part.

3. Think a bit about the last one, what happens if $\displaystyle f(x)=k>0$ where x is irrational?
What if you pick $\displaystyle \varepsilon$ such that $\displaystyle 0<\varepsilon<k$?

Know that every open interval contains irrational and rational numbers.

Try to visualize it.

4. Originally Posted by Plato
In the definition of continunity let $\displaystyle \varepsilon = \frac{{f(b) - \alpha }}{2} > 0$.

Now the way this forum works, you should show some effort on your part.
I'm still not seeing what I should be doing. I substituted it into the definition and tried tinkering around, but I'm not seeing anything.

5. Try this:
Define a function $\displaystyle g(x)=f(x)-\alpha$. g is continuous, because f(x) is and $\displaystyle \alpha$ is a constant.

Now $\displaystyle g(b)>0$, and if g is continuous then there must be an interval $\displaystyle I$ about b such that for every $\displaystyle x\in I$ , $\displaystyle g(x)>0$.

Is it easier to see now ?

6. Originally Posted by hjortur
Think a bit about the last one, what happens if $\displaystyle f(x)=k>0$ where x is irrational?
What if you pick $\displaystyle \varepsilon$ such that $\displaystyle 0<\varepsilon<k$?

Know that every open interval contains irrational and rational numbers.

Try to visualize it.
I don't see how being rational or irrational has anything to do with this problem!

7. Suppose that f:R-->R is continuous for all x and let f(q)=0 for all rationals q. Prove that f(x)=0 for all x
Then please let me know how you solve that without referring to rational or irrational numbers.