Question about the image of a continuous function not bounded from above or below

Question

I am stuck on the following problem:

Let $f :\mathbb R \to \mathbb R$ be a function such that $f$ is not bounded above nor bounded below. Show that if $f$ is continuous on $\mathbb R$ then the image of $f$ is $\mathbb R$.

I'm not sure how to exactly approach this problem, especially because it's so intuitive. I know you have to use the Intermediate Value Theorem.

Answer 1

Hint: for each $M>0$, there exist $x_1, x_2$ such that $f(x_1)>M$ and $f(x_2) < -M$. What does that tell you about $[-M, M]$?

Answer 2

Note that for every natural number $n$ $f$ must attain a value $y_n>n$ since otherwise $f$ would be bounded. By the IVT $f$ attains each value between $y_n$ and $y_{n+1}$ for any $n$. So $f$ is surjective onto $[m,\infty)$, where $m=\min_n y_n$. Doing the same for negative values and filling the gap in between, we conclude that $f$ has range $\mathbb R$.

Answer 3

Use IVT : Let's assume a real number $M$ such that $f(x)\neq M$ for all $x$, but by the unboundedness property of the function there is $x_1$ with $f(x_1)>M$ and $f(x_2)<M$, now using IVT $f$ is gonna attain all the values between $f(x_1), f(x_2)$, so it would attain $M$ too since $f(x_2)<M<f(x_1)$. So the assumtion was not true.