# Thread: Find the number of extremum

1. ## 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).

2. Both Captain Black and Noncomalg are correct.

I used the 2nd derivative test incorrectly.

3. Is x=2 the only extremum

4. 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

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