I always understood that $f'(c) >0 $ at a point $x=c$ means that there exists a $\delta>0$ so that f is monotonically increasing in the interval $(c-\delta,c+\delta)$.

However in the book A basic course in real analysis they have given an example of the following form $f(x)=x+2x^2sin(1/x)$ for $x \ne 0$ and $f(0)=0$ they have asked to prove that $f'(0)=1$ but $f$ is not monotonic in any interval around $0$.

This example has made me even more confused. Can someone explain what does $f'(c)>0$ at the point $x=c$ even mean?

  • $\begingroup$ See also math.stackexchange.com/q/4179995/442 and in particular the example $$\begin{cases}x+2x^2\sin(1/x) & x\neq\ 0\\ 0 & x=\ 0 \\ \end{cases}$$ $\endgroup$
    – GEdgar
    Sep 24, 2022 at 9:50
  • $\begingroup$ Alternative approach: go back to the basic definition. $f'(x) > 0 ~: ~x = c$ signifies that $$\lim_{h\to 0} \frac{f(c+h) - f(c)}{h}$$ exists and is positive. $\endgroup$ Sep 24, 2022 at 10:18
  • $\begingroup$ Re previous comment, what this signifies is that there exists some $L > 0$ such that $\forall \epsilon > 0, ~\exists \delta > 0,$ such that $$0 < |h| < \delta \implies \left| ~\frac{f(c+h) - f(c)}{h} - L ~\right| < \epsilon.$$ $\endgroup$ Sep 24, 2022 at 10:27

1 Answer 1


The fundamental property of the derivative is that for $x$ close to $c$, we have $$f(x)\approx f(c)+(x-c)f'(c)$$(the right-hand side is the standard formula for the line that goes through $(c,f(c))$, just as $f$ does, and has slope $f'(c)$).

The exact meaning of $\approx$ can of course be expanded upon (there is a rigorous definition for how good this approximation must be) but those are details for when calculations are needed or proofs are to be written. For an introductory, intuitive understanding of what the derivative is, this is the most important takeaway.

The definition of the derivative of $f$ at $c$ says nothing about miniscule oscillations of $f$ close to $c$, which is what your teacher's example illustrates. It only says that if such oscillations happen, they still won't take you far from the line $y=f(c)+(x-c)f'(c)$.

Now, if $f'(c)>0$, that just means that this line is increasing. It does, for instance, imply that there is a $\delta>0$ such that $f(x)<f(c)$ on $(c-\delta,c)$ and $f(x)>f(c)$ on $(c,c+\delta)$. But fundamentally, using the derivative at $c$ only lets you compare $f(x)$ to $f(c)$ and says nothing at all directly about how $f(x_1)$ compares to $f(x_2)$, even if $x_1$ and $x_2$ are both close to $c$.

  • 1
    $\begingroup$ typo: it is $f(x) \simeq f(c) + (x-c)f'(c)$! $\endgroup$
    – Didier
    Sep 24, 2022 at 7:39
  • $\begingroup$ @Didier Typo, brainfart, oversight. One of those, at any rate. Thank you. $\endgroup$
    – Arthur
    Sep 24, 2022 at 7:41

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