(Sorry if this question, along with some other recent questions, is trivial or there's errors in how I worded it. I'm a beginner calculus student)

Let $f:[a,b] \rightarrow \mathbb{R}$ be a continuous function twice-differentiable on the open interval $(a,b)$.

Proposition: if $c \in (a,b)$ is a strict local minimum of $f$, there exists a $\delta >0$ such that $f'$ is monotonically increasing in $(c, c + \delta)$.

Question 1: does this proposition hold?

Question 2: Let us consider a new function $g$ defined similarly to how $f$ was. Let $x_0$ satisfy $g'(x_0)=0$ and $g''(x_0)>0$. Moreover, there exists no $\delta >0$ such that $g'$ is monotonically increasing in $(x_0, x_0 + \delta)$. Then per contraposition on our original proposition, $x_0$ is not a local minimum of $g$. But this contradicts the 2nd derivative test. What's going on?

Note for $g$ to satisfy what we want, $g'$ could be some pathological differentiable function . Probably something you get after playing around with $\sin(\frac{1}{x})$; $g$ would of course be its primitive, surely not something expressible in elementary terms.

  • $\begingroup$ Hi, if around $0$, $g(x) = g(0) + a x^2 + o(x^2)$ with $a>0$ then $0$ is a local minimum. so you have to choose a function for which the 2nd derivative is not continuous such that the Taylor expansion of order $2$ doesn't hold. $\endgroup$
    – reuns
    Jan 9, 2016 at 7:52

1 Answer 1


Your proposition does not hold. As an example, define $f$ by $f(0) = 0$ and $$ f(x) = x^4 ( 2 + \sin \frac 1x) $$ for $x \ne 0$. $f$ is differentiable on $\Bbb R$ and has a minimum at $x = 0$.

The derivative is given by $f'(0) = 0$ and $$ f'(x) = 4x^3 ( 2 + \sin \frac 1x) - x^2 \cos \frac 1x = x^2 \bigl( 4x ( 2 + \sin \frac 1x) - \cos \frac 1x \bigr) $$ for $x \ne 0$.

$f'$ is also differentiable on $\Bbb R$ so that $f$ is twice differentiable. But $f'$ is not monotonically increasing in any interval $(0, \delta)$ because it takes both positive and negative values arbitrary close to $x=0$.

It does not even hold if $f$ has derivatives of all order. As an example, define $f$ by $f(0) = 0$, $$ f(x) = e^{-\frac 1x} ( 2 + \sin \frac {1}{x^2}) $$ for $x > 0$ and $f(-x) = f(x)$. $f$ has a minimum at $x = 0$ and has derivatives of all order (roughly because $e^{-\frac 1x}$ converges to zero for $x \to 0^{+}$ faster than any power of $x$).

For $x > 0$, $$ f'(x) = \frac{1}{x^3 }e^{-\frac 1x} \bigl( x( 2 + \sin \frac {1}{x^2}) - 2 \cos \frac {1}{x^2} \bigr) $$ which again takes both positive and negative values arbitrarily close to $x = 0$.

  • $\begingroup$ I'm curious, why did you delete the original answer?! $\endgroup$ Jan 9, 2016 at 7:56
  • $\begingroup$ @MathematicsStudent1122: Because it did not satisfy the requirement that $f''$ exists. I hope that is fixed now. $\endgroup$
    – Martin R
    Jan 9, 2016 at 7:57
  • $\begingroup$ That's helpful! What if $f$ is infinitely differentiable? Is there a lowest differentiability class for the statement to hold? Or is it always untrue? $\endgroup$ Jan 9, 2016 at 7:57
  • 1
    $\begingroup$ @MathematicsStudent1122: It is generally false. Something like $f(x) = \exp(-1/x) (2+\sin(1/x))$ for $x > 0$ should work as a counter-example. $\endgroup$
    – Martin R
    Jan 9, 2016 at 8:06
  • $\begingroup$ Okay, thanks!!! $\endgroup$ Jan 9, 2016 at 8:09

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