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  1. If a function $f(x)$ is continuous and increasing at point $x=a,$ then there is a nbhd $(x-\delta,x+\delta),\delta>0$ where the function is also increasing.
  2. if $f' (x_0)$ is positive, then for $x$ nearby but smaller than $x_0$ the values $f(x)$ will be less than $f(x_0)$, but for $x$ nearby but larger than $x_0$, the values of $f(x)$ will be larger than $f(x_0)$. This says something like $f$ is an increasing function near $x_0$, but not quite.
Styles
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    What does it mean for a function to be increasing at a point ? – Kolja Aug 10 '18 at 21:18
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    @Kolja A function $f$ is increasing at $x$ if $f(t)>f(x)$ for every $t>x$ close enough to $x$ and $f(t)<f(x)$ for every $t<x$ close enough to $x$. See https://math.stackexchange.com/questions/364576/can-a-function-be-increasing-at-a-point – mfl Aug 10 '18 at 21:31
  • @mfl The question is not whether there is a definition (we know there is one), but which one OP wants to use. – M. Winter Aug 11 '18 at 07:56

2 Answers2

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The two statements are not the same. One assumes differentibity and the other does not.

Mohammad Riazi-Kermani
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  • The question is: is this relevant? Are "differentiable functions are continuous" and "strictly increasing functions are injective" equivalent? Both are true, hence one is true whenever the other one is. But only one is about differentiability. Maybe "being the same" means talking about the same aspects rather than being equivalent? It is not clear (from the question). – M. Winter Aug 11 '18 at 08:27
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Part 1.-

Consider $f(x)=x\left(\sin \frac1x +2\right), x>0, f(0)=0.$ It is clear that for $x>0$ it is $f(x)>0.$ But $$f'(x)=\sin\frac1x+2-\frac1x\cos\frac1x.$$ So $f$ is increasing at $0$ but not increasing in $(0,\delta).$ (Of course, you can modify the function to work on $(-\delta,\delta).$

Edit

An example can be found in the nice book (see Problem 1.13) https://stemtec.aut.ac.nz/__data/assets/pdf_file/0003/57639/Counterexamples-in-Calculus-MAA-e-book.pdf

Part 2.-

If $f'(x_0)=\lim_{x\to x_0}\dfrac{f(x)-f(x_0)}{x-x_0}>0$ we have that there exists $\delta>0$ such that $x\in (x_0-\delta,x_0)\implies f(x)<f(x_0)$ and $x\in (x_0,x_0+\delta)\implies f(x)>f(x_0).$ Thus $f$ is increasing at $x_0.$

Conclusion

Both statements are not the same. The first one is false while the second one holds.

mfl
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