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Question: Let $f$ be defined on an open interval $(a, b)$ and assume that for some $c \in (a, b)$ $f^{\prime} (c)> 0$ then show that there is a open ball around $c$, say $B(c) \subseteq (a, b)$ in which $f(x)> f(c)$ if $x>c$ and $f(x)< f(c)$ if $x<c$. Note: $f^{\prime}$ denotes the derivative of the function $f$.

What I have understood is that I have to show that function $f$ is strictly increasing in the interval $(a, b)$. Given that $f^{\prime} (c)> 0$ which implies that $f$ must be differentiable. Now, I am trying to apply the Lagrange mean value theorem. But I am not sure which interval to take so that I can prove desired inequality. Also, how can I introduce open ball around $c$ so that $B(c) \subseteq (a, b)$ and desired inequality holds true there.

Thank you

Jonas Meyer
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Srijan
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1 Answers1

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No, $f$ need not be increasing in any interval surrounding $c$. The result here follows directly from the definition of the derivative at $c$.

$\lim\limits_{x\to c}\dfrac{f(x)-f(c)}{x-c} =f'(c)>0$,

which implies that when $x$ is sufficiently close to $c$, i.e. for $x$ in some interval $(c-\delta,c+\delta)$, $\dfrac{f(x)-f(c)}{x-c}$ is positive. For $x>c$, this implies $f(x)>f(c)$. For $x<c$, this implies $f(x)<f(c)$.

To see why you can't show $f$ is increasing in an interval, see Differentiable+Not monotone. Even though the link doesn't work in the answer, hopefully the reference given or the comments or googling will lead you to more info if you're interested.

For just a particular point like you're asking about, $f(x) = x+2x^2\sin(1/x)$ (and $f(0)=0$) provides an example where $f$ is not monotone in any interval containing $0$, although $f'(0)=1$. When $x\neq 0$ we have $f'(x)=1+4x\sin(1/x)-2\cos(1/x)$, which is negative arbitrarily close to $0$, e.g. when $x=\dfrac{1}{2\pi n}$ for each nonzero integer $n$. By the same reasoning as in the solution to your problem, this implies that $f$ can't be increasing in any interval containing a $\dfrac{1}{2n\pi}$, while every interval containing $0$ has infinitely many of these. Without using derivatives, you can verify that when $n$ is sufficiently large, $f\left(\dfrac{1}{2n\pi}\right)-f\left(\dfrac1{2n\pi+\frac\pi2}\right)<0$.

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Jonas Meyer
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  • Thank you for very clear answer. I fully understood. I have just one doubt. What is the difference between this condition $f(x)> f(c)$ if $x>c$, around certain ball at $c$ and definition of monotonic function? Thank you again. – Srijan Feb 16 '17 at 05:27
  • It means the values to the right are bigger than $f(c)$, but they don't stay in order relative to each other. You can have $c<x<y<c+\delta$ where $\delta $ is small enough to ensure that $f(x)>f(c)$ and $f(y)>c$, even though $f(x)>f(y)$, and there are examples where this can happen no matter how small $\delta$ is. The other question is for really dramatic cases of this. I might add a concrete example for just a particular $c$. – Jonas Meyer Feb 16 '17 at 05:41
  • Thank you very much for your patience and an elegant answer. As for now, I am not able to construct such an example but your statement is very convincing to me. Thank you again. – Srijan Feb 16 '17 at 05:47
  • @srijan: I added an example. – Jonas Meyer Feb 16 '17 at 05:52
  • I am clear now. At one point, I think you mean $f^{\prime}(x) = 1+4x\sin(1/x)-2\cos(1/x)$ is non negative$? Am I right?. A sincere thanks from my side dear professor. – Srijan Feb 16 '17 at 05:54
  • @srijan: No, I mean that arbitarily close to $0$ there are values of $x$ such that $f'(x)<0$, and then specifically note that you can take $x=1/(2n\pi)$ to see this. This is true even though $f'(0)=1$. You're welcome. – Jonas Meyer Feb 16 '17 at 06:01
  • Thank you got your point. I didn,t pay attention to $x$. No more confusion. :) – Srijan Feb 16 '17 at 06:03