# A continuous function having an inverse <=> conditions on a derivative?

1. Jan 16, 2013

### hb1547

Sorry for the poorly-worded title.

I help tutor kids with pre-calculus, and they're working inverse functions now. They use the "horizontal line test" to see if a function will have an inverse or not by seeing visually if it's one-to-one.

I was thinking about what that might imply. If a function passes the horizontal line test (let's assume it's domain is ℝ and that it's continuous everywhere), then it must not have any extremum, since then it'd fail the test. So this would imply that:
$$f'(x) \le 0 \hspace{3pt}\mathrm{or }\hspace{3pt} f'(x) \ge 0 \hspace{3pt} \forall x$$
I'm wondering if this is always true, and so if it's possible to prove that a bijective function must have a derivative that's either zero or positive/negative everywhere, and I'm wondering if the converse is also true.

I'm more of a physics-guy than a math-guy, but I do find math interesting and these are the types of questions that I like to think about. Any thoughts?

2. Jan 16, 2013

### Simon Bridge

I would imagine that it is always true for the kinds of functions and inverses you are considering and proving the horizontal line test would be sufficient proof of this too. I think the converse logically follows by completeness - to have an inverse, the derivative of the bilinear form must be definite.

3. Jan 17, 2013

### THSMathWhiz

Consider the function
$f(x)=\begin{cases}(x+2)^3,&x<-2,\\0,&-2\leq x\leq 2,\\(x-2)^3,&x>2.\end{cases}$
You can see that $f(x)$ is continuous on $\mathbb{R}$ and that $f'(x)\geq0\forall x\in\mathbb{R}.$ However, this function fails the horizontal line test on $[-2,2]$ and therefore does not have a continuous inverse. A sufficient condition for the converse to be true would be a strict inequality.