1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proving f(x) has one real root

  1. Dec 19, 2011 #1
    The problem asked to prove by contradiction that the function has only one real root. They used Roll'es theorem and proved that f ' (x) =/= 0 then they concluded that because f ' (x) =/= 0 then by contradiction the function can have atleast one real root...


    Look at the picture it has the answer to the solution...

    Question:
    I dont understand the logic... I made up a graph where f ' (x) = 0 but yet f(x) still only has one real root.... so how is disproving that f'(x) can never equal zero prove that f(x) has atleast only one real root when I obviously showed a scenario in which f'= 0 but f only has one root
     

    Attached Files:

    • math.jpg
      math.jpg
      File size:
      23.5 KB
      Views:
      150
  2. jcsd
  3. Dec 19, 2011 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    The proof is saying because f'(x)>0 for all x, there is only one real root. You're confusing that statement with its converse, "f(x) has only one real root; therefore, f'(x)>0," which isn't generally true.
     
  4. Dec 19, 2011 #3
    thank you
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook