# Thread: Continuous Functions

1. ## Continuous Functions

I'm having some trouble with this question:

$Let f : [a,b] \rightarrow \mathbb{R}$ $be\ a\ continuous\ function\ such\ that\ f(x)>0$ $for\ all\ x \in [a,b].$ $Show\ that\ there\ is\ a\ positive$ $constant\ c\ such\ that\ f(x) \geq c$ $for\ all\ x \in [a,b].$ $Give\ examples\ to\ show\ that\ the\ conclusion\ fails$ $if\ [a,b]\ is\ replaced\ by\ (0,1] or [0,\inf ).$

2. This is just the extreme value theorem. Since f is continuous and positive on a compact set [a,b], it achieves a positive minimum in [a,b].

An example of a function for which it fails is $f(x) = \sqrt{x}$ on $(0,1]$ or $f(x) = {1\over 1+ x}$ on $[0,+\infty)$.

3. Originally Posted by acevipa
I'm having some trouble with this question:

$Let f : [a,b] \rightarrow \mathbb{R}$ $be\ a\ continuous\ function\ such\ that\ f(x)>0$ $for\ all\ x \in [a,b].$ $Show\ that\ there\ is\ a\ positive$ $constant\ c\ such\ that\ f(x) \geq c$ $for\ all\ x \in [a,b].$ $Give\ examples\ to\ show\ that\ the\ conclusion\ fails$ $if\ [a,b]\ is\ replaced\ by\ (0,1] or [0,\inf ).$
How you go about proving this much depends on what you already know about compact intervalls like [a;b], and continuous functions.
For example, there is a theorem that says that for a continuous function on a compact interval [a;b], there are $x_{\text{min}},x_{\text{max}}\in[a;b]$, such that $f(x_{\text{min}})\leq f(x)\leq f(x_{\text{max}})$, for all $x\in [a;b]$.
So, if you can use that theorem, you can just put $c := f(x_{\text{min}})$, but if you cannot use that theorem, you have to (essentially) prove it.