January 30th, 2016, 08:59 AM  #1 
Member Joined: Apr 2013 Posts: 36 Thanks: 0  stuck!
Hello! I'm stuck with the following problem... maybe someone can give me ideas on how to go about it... I have two functions f(x) and g(x) and I know that f has one zero at y. I need to show that g(y) > 0. Any suggestions on what to consider in order to prove this? I know the result should be true. Thanks! Natalie Last edited by skipjack; January 30th, 2016 at 09:33 AM. 
January 30th, 2016, 09:04 AM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,472 Thanks: 884 Math Focus: Elementary mathematics and beyond 
What's the relationship between f(x) and g(x)?

January 30th, 2016, 09:11 AM  #3 
Member Joined: Apr 2013 Posts: 36 Thanks: 0 
What type of relation ?

January 30th, 2016, 09:14 AM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,472 Thanks: 884 Math Focus: Elementary mathematics and beyond 
Your original post appears incomplete. f(y) = 0 does not imply g(y) > 0 unless there is some relationship between f and g.

January 30th, 2016, 09:41 AM  #5 
Member Joined: Apr 2013 Posts: 36 Thanks: 0 
yes yes they are related of course apologies for that...the expressions look like this... the constants $\displaystyle M_i, P_i , i=1,2,...6$ are related as well...I just need ideas on what to consider to prove this result...it is not true in general but for the constants i have it is. for fixed A > 0: $\displaystyle f(y;A) = A*[M_1 y^{\gamma_1+\gamma_3} + M_2 y^{\gamma_1+\gamma_2} + M_3 y^{\gamma_2+\gamma_3} + M_4 y^{\gamma_1 + 1} + M_5 y^{\gamma_2 + 1} + M_6 y^{\gamma_3 + 1}]  [P_1 y^{\gamma_1+\gamma_3} + P_2 y^{\gamma_1+\gamma_2} + P_3 y^{\gamma_2+\gamma_3} + P_4 y^{\gamma_1 } + P_5 y^{\gamma_2} + P_6 y^{\gamma_3 }] $ I know that for each fixed $\displaystyle A > 0, y \mapsto f(y;A) $ has a unique root. $\displaystyle g(y;A) = A*[M_1 (\gamma_1 + \gamma_3 1) y^{\gamma_1+\gamma_3} + M_2 (\gamma_1 + \gamma_2 1) y^{\gamma_1+\gamma_2} + M_3 (\gamma_2 + \gamma_3 1)y^{\gamma_2+\gamma_3} + M_4 \gamma_1 y^{\gamma_1 + 1} + M_5 \gamma_2 y^{\gamma_2 + 1} + M_6 \gamma_3 y^{\gamma_3 + 1}]  [P_1 (\gamma_1 + \gamma_3 1)y^{\gamma_1+\gamma_3} + P_2 (\gamma_1 + \gamma_2  1)y^{\gamma_1+\gamma_2} + P_3 (\gamma_2 + \gamma_3 1)y^{\gamma_2+\gamma_3} + P_4 (\gamma_1 1) y^{\gamma_1} + P_5 (\gamma_2  1) y^{\gamma_2 } + P_6 (\gamma_3 1 ) y^{\gamma_3 }] $ 

Tags 
stuck 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
I'm stuck!  Isabel12144  Probability and Statistics  4  February 19th, 2015 03:25 PM 
I am so stuck with this !  Ole Daniel  Algebra  4  November 22nd, 2014 03:53 PM 
IM STUCK PLZ HELP...  khalidd  Algebra  1  September 9th, 2010 11:19 AM 
So now you are stuck with me  MathyMattyCalls  New Users  3  April 22nd, 2010 11:22 AM 
Please help i am stuck  valley_girl1919  Calculus  1  October 6th, 2009 04:17 AM 