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

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A continuous function that passes the horizontal line test is one-to-one and must not have any extrema, implying its derivative is either non-negative or non-positive across its domain. The discussion explores whether this condition guarantees that a bijective function has a derivative that is consistently zero or strictly positive/negative everywhere. It is suggested that proving the horizontal line test could serve as sufficient proof for this relationship. An example function is provided, demonstrating that while it is continuous and has a non-negative derivative, it fails the horizontal line test and thus lacks a continuous inverse. The conversation concludes that a strict inequality in the derivative may be necessary for the converse to hold true.
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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&#039;(x) \le 0 \hspace{3pt}\mathrm{or }\hspace{3pt} f&#039;(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?
 
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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
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.
 
Simon Bridge said:
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.

Consider the function
f(x)=\begin{cases}(x+2)^3,&amp;x&lt;-2,\\0,&amp;-2\leq x\leq 2,\\(x-2)^3,&amp;x&gt;2.\end{cases}
You can see that f(x) is continuous on \mathbb{R} and that f&#039;(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.
 
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

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