Consider the function $\displaystyle f:R\rightarrow R$

$\displaystyle

f(x)=(x-1)(x-2)^2(x-3)^3(x-4)^4(x-5)^5......(x-n)^n

$

Find the number of points at which the function has an extremum(i.e points of local maxima/minima).

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- Oct 29th 2009, 05:39 PMpankajFind the number of extremum
Consider the function $\displaystyle f:R\rightarrow R$

$\displaystyle

f(x)=(x-1)(x-2)^2(x-3)^3(x-4)^4(x-5)^5......(x-n)^n

$

Find the number of points at which the function has an extremum(i.e points of local maxima/minima). - Oct 29th 2009, 06:47 PMTheEmptySet
Both Captain Black and Noncomalg are correct.

I used the 2nd derivative test incorrectly. - Oct 29th 2009, 09:25 PMpankaj
Is x=2 the only extremum

- Oct 30th 2009, 12:42 AMCaptainBlack
- Oct 30th 2009, 02:13 AMNonCommAlg
considering the fact that the graph of a polynomial at the roots with even (odd) multiplicity touches (crosses) the x-axis, and hoping that nothing weird happens in between, i believe that the

number of required points is: $\displaystyle n + \left \lfloor \frac{n}{2} \right \rfloor -1.$