How do I show that a derivative of a polynomial has a zero in an interval?

Charles49
Messages
87
Reaction score
0

Homework Statement



If an even degree polynomial of order 2n intersects the x-axis twice, how do I show that the (2n-1) th derivative has a zero in that interval?


Homework Equations



example: let g(x)=x^3(1-x). Show without computation that g'''(c) =0 for some c in (0, 1).

The Attempt at a Solution



I know that g' has a zero at x=1/2 because that is a maximum. I don't know how I can be sure that the 2nd and 3rd derivative have a zero in this interval.
 
Physics news on Phys.org
consider what happens to terms less of order (2n-2) and less...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top