Results 1 to 5 of 5

Math Help - Using rolle's theorem to show a function has only one real root

  1. #1
    Member
    Joined
    Oct 2009
    Posts
    79

    Using rolle's theorem to show a function has only one real root

    \longrightarrow Use rolle's theorem to show that f(x)=3x-cosx-1 has only one real root.

    So, I know you use IVT to say it has at least one root and then contradict it with Rolle's, but in order to use the IVT you have to pick two points- one negative and one positive so there is point where it was zero. Is there any method for this or do you pick numbers and check to see if it works?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by hazecraze View Post
    \longrightarrow Use rolle's theorem to show that f(x)=3x-cosx-1 has only one real root.

    So, I know you use IVT to say it has at least one root and then contradict it with Rolle's, but in order to use the IVT you have to pick two points- one negative and one positive so there is point where it was zero. Is there any method for this or do you pick numbers and check to see if it works?

    1) Prove your function is always increasing (hint: check f'(x))

    2) Check that [tex]f(0)f(1)<0[\math]

    3) Now use the IVT and solve your problem.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2009
    Posts
    79
    I'm really just concerned about finding the interval where f(c)=0.

    I got stuck trying to find where it is increasing:
    3 + sin(x)=0
    sin(x)=-3
    =

    But it seems (how the book writes it) that you should just pick small numbers like 0 and 1 to see if it works. I guess in this case: 0 and pi works-(f(0) is negative and f(pi) is positive).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by hazecraze View Post
    I'm really just concerned about finding the interval where f(c)=0.

    I got stuck trying to find where it is increasing:
    3 + sin(x)=0
    sin(x)=-3
    =

    But it seems (how the book writes it) that you should just pick small numbers like 0 and 1 to see if it works. I guess in this case: 0 and pi works-(f(0) is negative and f(pi) is positive).

    If you're doing calculus then you must know some basic trigonometry, don't you? Well, as |\sin x|\leq 1\,\,\forall\,\,x\in \mathbb{R}, then f'(x)>0 ALWAYS...right? And thus your functions is always strictly increasing and then it has at most one zero.

    Now check point (2) in my first message and solve your problem.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2009
    Posts
    151
    There is no reason to use IVT, Rolle's theorem alone does the trick.

    Assume that f(x)=3x-cosx-1 has at least two roots x=a and x=b.
    Then f(a)=f(b)=0.

    Now by Rolle's theorem there excists a c such that f'(c)=0.

    But what is f'(x) ?

    Hope that helps.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Show that polynomial has one real root
    Posted in the Calculus Forum
    Replies: 6
    Last Post: September 8th 2010, 12:08 PM
  2. Replies: 1
    Last Post: February 3rd 2010, 02:44 PM
  3. Replies: 3
    Last Post: September 20th 2009, 12:37 PM
  4. show equation has exactly one real root
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 10th 2008, 09:29 PM
  5. Replies: 3
    Last Post: March 13th 2008, 08:26 AM

Search Tags


/mathhelpforum @mathhelpforum