Author |
Topic: g(x) = g'(x) (Read 473 times) |
|
ThudnBlunder
wu::riddles Moderator
Uberpuzzler
The dewdrop slides into the shining Sea
Gender: 
Posts: 4489
|
 |
g(x) = g'(x)
« on: Jan 20th, 2009, 6:02pm » |
 Quote  Modify
|
If g(x) is a polynomial with a real root, show that there always exists an x such that g(x) = g'(x).
|
|
 IP Logged |
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
|
|
|
teekyman
Full Member
Gender:
Posts: 199
|
|
Re: g(x) = g'(x)
« Reply #1 on: Jan 21st, 2009, 12:57am » |
Quote Modify
|
wlog assume the leading coefficient of g(x) is positive.
1. if g(x) is an odd polynomial, then g(x) - g'(x) is also odd and has at least one real root
2. If g(x) is even and has a root tangent to the x axis, then that root satisfies g(x) = g'(x)
3. Otherwise, g(x) is even and must have at least two nonrepeating roots. Say the roots are, in increasing order, x1, x2, ... xk. we know g'(x1) is negative, g'(xn) is positive, thus g(x) - g'(x) is positive at x1, negative at xn, thus by IMT it must be 0 at some point.
|
| « Last Edit: Jan 21st, 2009, 1:00am by teekyman » |
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print