Question 1041435
<font face="Times New Roman" size="+2">


A2. Suppose that *[tex \Large f] is a polynomial of degree *[tex \Large n], where *[tex \Large n\ \geq\ 2]. Show that if *[tex \Large f ] has *[tex \Large n] distinct roots, then *[tex \Large f'(x)] must have *[tex \Large n\ -\ 1] distinct roots.


Between any two roots of *[tex \Large f(x)] there must be a root of *[tex \Large f'(x)] by Rolle's Theorem.  But since all *[tex \Large n] roots of *[tex \Large f(x)] are distinct and since *[tex \Large f'(x)] is perforce of degree *[tex \Large n\ -\ 1], there can only be *[tex \Large n\ -\ 1] roots of *[tex \Large f'(x)] and they must be distinct. 


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

</font>