Find the number of extremum

• Oct 29th 2009, 05:39 PM
pankaj
Find 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 PM
TheEmptySet
Both Captain Black and Noncomalg are correct.

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

Originally Posted by TheEmptySet

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

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

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

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

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

Originally Posted by pankaj
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).

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