[passer-by]

1. We'll prove that $ f(x)= 0 $ for all irrational numbers

Let $x$ be an irrational number

Since Q is dense in R , we can find sequence $ (x_n) \subset Q $ converges to irrational $x$

And by continuity of $f$ , then we have $ f(x_n) $ converges to $ f(x)$

But from definition of $f$ , we have $ f(x_n)= 0 $ for all $n$

Hence, $ f(x)=0 $ for all irrational $x$

NOTE: This problem leads to interesting corollary: If f,g are continuous on R and f(r)=g(r) for all rational numbers r ,then f(x)=g(x) for all real numbers x


2. Clearly, such $f$ is continuous on R

Since $f(1) < 0$ and $f(2) >0$ then ,by intermediate value theorem, there exists $ r \in (1,2) $ such that $ f(r)=0$

Use this theorem again on $(-8,0)$ since $f(-8) >0$ and $f(0)<0$


3. Define $ g(x)= f(x) - \beta $ Then g is continuous on R and $g(x_0) <0 $

By definition of continuous function, there exists a neighborhood of $x_0$ such that

$ \mid g(x)-g(x_0) \mid < \frac{-g(x_0)}{2} $

Equivalently, $ g(x) < \frac{g(x_0)}{2} < 0$

So $ f(x) < \beta $ in a neighborhood of $x_0$

4. (Like proof for the case f,g are continuous)

5. Not always.

For example, $ f(x)=\left\{\begin{array} &-1& x \leq 0 \\ x-1& x>0 \end{array}\right. $
$ g(x)=\left\{\begin{array} &x& x \leq 0 \\ 0& x>0 \end{array}\right. $

But $ (fg)(x)=\left\{\begin{array} &-x& x \leq 0 \\ 0& x>0 \end{array}\right. $ which is decreasing function.