Finding continuous f where f is not bounded .

I am finding a function f which is continuous on a closed interval but not bounded on the interval.

• You will not find one. Commented May 4, 2013 at 6:17

Isn't $[0, \infty)$ a closed interval? Then just take $x$ in that case.

As André stated above, you will not find such a function.

On the one hand, think about what unboundedness on a finite interval means: it means arbitrarily large jumps in a finite interval. No such function can be continuous.

More precisely, there is the Extreme Value Theorem.

Well to formalize it a bit, what you're saying is

continuous on a closed interval but not bounded on the interval

The not bounded on closed interval means for some interval $[a,b]$:

$\forall M>0\in\mathbb{R} ,\ \ \exists x_0 \in [a,b]$ so that $f\left(x_0\right)>M$

But as mixedmath stated according to Extreme Value Theorem:

f is continuous in the closed and bounded interval [a,b], then f must attain its maximum and minimum value, each at least once. That is, there exist numbers $c$ and $d$ in $[a,b]$ such that $\forall x\in [a,b] , f(c)\leq f(x)\leq f(d)$

(courtesy of Wikipedia)

So now we can easily get a contradiction: let's just pick $M=2f(d)$ and try to use the first statement clearly it's a contradiction to the Extreme Value Theorem.

But as blitzer noted correctly it does depend on whether you have to have finite numbers as endpoints. Please read here more on intervals.