A while back a person posted this question on this site. I played with it for a while until I found (in my opinion) some pretty nice examples. A couple of minutes after I posted my solution the OP deleted the post, so I decided to ask-and-answer the question myself to share the examples.


(a) A function with exactly two horizontal asymptotes and exactly two vertical asymptotes, but is defined everywhere else on $\mathbb{R}$.

(b) A continuous function on $\mathbb{R}$ with $f(2)=−3$, $f(−3)=2$, and whose graph has no 𝑥-intercept.

(c) An everywhere continuous odd function on $\mathbb{R}$ which achieves its absolute maximum value at exactly two points.


1 Answer 1


a) A function that works is $$ \frac{1}{x(x-1)} +\arctan(x) $$ since the first part gives you the vertical asymptotes and the $\arctan(x)$ part gives you the horizontal ones.

Example a)

b) It's impossible. Notice that at $x = -3$ you are above the $x$-axis (at a height of $2$) and that at $x=2$ you are below the $x$-axis (at a height of $-3$). If the function is continuous you should be able to draw its graph without lifting your pencil. So can you draw a line from above the $x$-axis to below the $x$-axis without lifting your pencil and without crossing the $x$-axis?

Bolzano's theorem (or the stronger Intermediate value theorem) proves the impossibility of this.

c) An example that is continuous (and even differentiable) is $$ f(x) = \begin{cases} e^{2\pi+x} -1, & \text{for }\ x \le -2\pi\\ \sin(x), & \text{for }\ -2\pi<x < 2\pi\\ -e^{2\pi-x} +1, & \text{for }\ x \ge 2\pi\\ \end{cases} $$ Example c)


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