Show that $f$ is bounded. Let $-\infty<a<b<\infty$. Suppose $f$ is continuous on $[a,b]$. Show that $f$ is bounded on $[a,b].$ 
We are supposed to use intermediate value theorem for this problem. But, I don't understand how to approach. All I can do is let, $$g(x) = f(x) - x$$  Then, $$g(a) = f(a)-a < 0$$ and $$g(b) = f(b) - b >0$$ And, according to I.V.T, $\exists c \in [a,b]$ s.t. $g(c ) = 0$. Now what? How can I connect all these? I am clueless here. Any help would be much appreciated. Thanks. 
 A: Assume a function $f(x)$ is not bounded on $[a,b]$. We prove it is not continuous.Then, for each $n \in \mathbb{N}$ we can find $(x_{n}) \in [a,b]$ such that $f(x_{n})>n$. Now extract a subsequence $x_{n_{k}}$ of $x_{n}$ converging to $x \in [a,b]$ (Bolzano Weirstrass and $[a,b]$ is closed). So $f(x_{n_{k}}) > n_{k}$. Now take $k \to \infty$. By comaraison, the sequence $(f(x_{n_{k}}))$ diverges. The sequential criterion for continuity is not satisfied!.
A: You could use the fact that continuous functions map compact sets onto compact sets.  Since $[a,b]$ is compact, $f([a,b])$ is compact and hence closed and bounded.
A: I am going to guess that you are mixing up related concepts. The proof that a continuous function $f$ is bounded on a compact interval $[a,b]$ most naturally uses the definition of continuity and the definition of compactness, and various properties of how compactness is preserved under continuous maps.
This has little to do with the intermediate value theorem (which is all about connectedness and continuous maps).
