Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

mukushla Group TitleBest ResponseYou've already chosen the best response.3
for \(x\le0\) its obvious that \(x^8 x^7 + x^2 x +15 >0\) u just prove for \(x>0\) : \(x^8 x^7 + x^2 x +15 >0\)
 one year ago

Aryang Group TitleBest ResponseYou've already chosen the best response.0
and does descartes' theorem help a great deal ? according to that,,it should have max 4 +ve roots..
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.3
no need to descartes...
 one year ago

Aryang Group TitleBest ResponseYou've already chosen the best response.0
well what then ?
 one year ago

Aryang Group TitleBest ResponseYou've already chosen the best response.0
ohh wait,,leme try..
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.3
well just show that \(x^8 x^7 + x^2 x +15>0\) for every real number
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
descartes helps a lot lol
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
http://mathworld.wolfram.com/DescartesSignRule.html
 one year ago

Aryang Group TitleBest ResponseYou've already chosen the best response.0
for x>1 ,, x^8 > x^7 and x^2 >x so f(x) > 0 for x>1 for 0<x<1 , x> x^2 > x^7 > x^8 we then have 15  (fraction + fraction) ,,which is obviously >0 so f(x) is always >0 am i right ?
 one year ago

Aryang Group TitleBest ResponseYou've already chosen the best response.0
ohh..thank you @mukushla your little hint only helped me a lot :)
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.