# Find the number of extremum

• October 29th 2009, 05:39 PM
pankaj
Find the number of extremum
Consider the function $f:R\rightarrow R$

$
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).
• October 29th 2009, 06:47 PM
TheEmptySet
Both Captain Black and Noncomalg are correct.

I used the 2nd derivative test incorrectly.
• October 29th 2009, 09:25 PM
pankaj
Is x=2 the only extremum
• October 30th 2009, 12:42 AM
CaptainBlack
Quote:

Originally Posted by TheEmptySet

Now for any $m >2$ notice thatwhen $x=m$ that $f'(m)=f''(m)=0$

So m is a saddle point for every $m>2$ and $x=2$ is either a max or a min.

Would your argument not mean that $y=x^4$ does not have a minima but only a saddle point at $x=0$?

(you have misinterpreted the second derivative test for classification of extrema)

CB
• October 30th 2009, 02:13 AM
NonCommAlg
Quote:

Originally Posted by pankaj
Consider the function $f:R\rightarrow R$

$
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).

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: $n + \left \lfloor \frac{n}{2} \right \rfloor -1.$