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If $f'(x)=(x-a)^{2n}(x-b)^{2m+1}$ where $m,n\in \mathbb{Z} $ then prove that there is neither a minimum nor a maximum at $a $,minimum at $b$ for $f(x)$ . Here I dont know whether $m>n,b>a $ thus even though if I calculate the value at $(2m+2)^{\text{th}}$ derivative I cant tell whether Its positive or negative thus I cant tell whether it minima or maxima . So I am basically stuck in taking a step further of Leibnitz rule.Any help is appreciated. Thanks!

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  • $\begingroup$ @ArchisWlankar regards. The maximum and minimum is for $f(x)$ or $f'(x)$? $\endgroup$ – Arief Anbiya Oct 27 '17 at 14:36
  • $\begingroup$ Sorry its for $f(x)$ $\endgroup$ – Archis Welankar Oct 27 '17 at 18:42
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The first non-zero derivatives at $x=a$ and $x=b$ are:

$$ f^{(2n+1)}(a) = (2n)!(a-b)^{2m+1} \\ f^{(2m+2)}(b) = (2m+1)!(b-a)^{2n} $$

To see this: at $x=a$, you repeatedly differentiate with the product rule, but notice that almost all the terms have a coefficient of $(x-a)$. The first time this isn't the case is for the term where you differentiate $(x-a)^{2n}$ until the $x$ goes away, i.e. after $2n$ differentiations. Its coefficient (from the product rule) is $(x-b)^{2m+1}$, which is non-zero. (To be explicit, all the terms where $(x-b)^{2m+1}$ is differentiated keep a coefficient of $(x-a)$ raised to some power, so all these are zero). A similar argument applies for $x=b$.

The derivative test tells us that if the first non-zero derivative is after an odd number of differentiations, then you have an inflection point (the case for $x=a$).

If the first non-zero derivative is after an even number of differentiations, then it's a minimum if that derivative is positive, and it is, since $(b-a)^{2n}$ is positive (the case for $x=b$).

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I may contribute. I presume that the maximum and minimum are with respect to $f(x)$.


We have $f'(x) = (x-a)^{2n} (x-b)^{2m+1} $.


The key is to notice the sign of $f'(x)$ (the monotonicity). Also that $(x-a)^{2n}$ is always positive, and $(x-b)^{2m+1}$ only positive when $x>b$.

  • Now if $a < b$ :

Then for $x < a$, we will have $f'(x) < 0$. For $ a<x<b $, we will have the same sign ($f'(x) < 0$). This means that $f(x)$ will be strictly decreasing in $(-\infty, b)-\{a\}$, with $a \in (-\infty, b)$. So $f(a)$ is not a peak ( neither a minimum or maximum).

And for $x > b$ we will have $f'(x) >0$. Combine this with the fact that $f'(x)<0$ for $a < x <b$, we can conclude that $f(b)$ will be a minimum value.

  • Now if $b<a$ :

Then for $x < b$, we will have $f'(x) < 0$. For $ b<x<a$, we will have the opposite sign ($f'(x) > 0$). This means that $f(b)$ will be a minimum value.

And for $x > a$ we will have $f'(x) >0$. Combine this with the fact that also $f'(x)>0$ for $b < x <a$, we can conclude that $f$ is strictly increasing in $(b, +\infty)-\{a\}$, with $a \in (b, + \infty)$. So $f(a)$ is not a peak ( neither a minimum or maximum).

Hope this helps.

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