Results 1 to 4 of 4

Math Help - intermediate value theorem question

  1. #1
    Newbie
    Joined
    Oct 2008
    From
    Toronto
    Posts
    11

    intermediate value theorem question

    I have a term test coming Nov 5th, so my posts will be here for the next couple weeks while I cram to complete my homework. Here goes:


    If f(x)=(x^2)+10sinx, show that there is a number c such that f(c)=1000

    This is what I have so far:

    make the function equal 1000, so,

    1000=(c^2)+10sin(c)

    I would like to solve for c now, but for some reason, the idea in the SSM says that the IVT applies here. I would like to just start plugging in trial numbers to see what values I can get slightly above and below 1000, but I'm not even too sure on how to evaluate (let's say) 10sin(20) without a calculator.

    How do I go about continuing to solve this not knowing how to evaluate the sin of my trial numbers without a calculator?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Jul 2008
    Posts
    46
    You're on the right track!

    what you need to do is to move the 1000 to the RHS to get an equation to equal 0.

    then let f(x)=(x^2)+10sinx-1000

    Then all you need to do is to find a value of x for when f(x)<0, and a value of x for when f(x)>0 (you can find "nice" values of x to make the sin function easy to calculate; try x=0 for starters) - proving that there is a solution between these two points, hence this is where the IVT comes in.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2008
    From
    Toronto
    Posts
    11

    Evaluating sin

    Hmm.. thank you but I'm still a little confused about evaluating the sin of integers.

    Sin functions tend to use radians, so would the values of c that are substituted in after I try 0 be pi values, or integers?

    Like, should I try integers, such as 2 next, then 3 and so on? How do I evaluate sin(2)?
    If c=2, then
    0=2^2+10*sin(2)-1000
    0=4+10*sin(2)-1000
    ((1000-4)/10)=sin(2)
    498/5 = sin(2)
    Nothing here tells me that I'm getting anywhere with the idea that there is an x that yields a y value of 1000.

    Or do I try pi then 2*pi and so on?
    If c=pi, then
    0 = pi^2 + 10sin(pi) - 1000
    0 = pi^2 + 10*(0) - 1000
    0 = pi^2 - 1000
    +-root(1000) = pi
    +-10 = pi

    I'm still waiting for that smile and sigh of relief... What am I doing wrong?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by krystaline86 View Post
    I have a term test coming Nov 5th, so my posts will be here for the next couple weeks while I cram to complete my homework. Here goes:


    If f(x)=(x^2)+10sinx, show that there is a number c such that f(c)=1000

    [snip]
    f(x) is continuous over [0, 100].

    f(0) = 0.

    f(100) = 10,000 + (a number between -10 and 10) > 1000.

    Therefore by the IVT there exists a number c such that 0 < c < 100 and f(c) = 1000.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Intermediate Value Theorem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 29th 2011, 01:04 PM
  2. Intermediate Value Theorem question
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 22nd 2009, 08:54 AM
  3. intermediate value theorem question
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 7th 2009, 11:46 AM
  4. Question about Intermediate Value Theorem on diff. Topologies
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: April 8th 2009, 04:37 PM
  5. intermediate value theorem/rolle's theorem
    Posted in the Calculus Forum
    Replies: 6
    Last Post: December 8th 2007, 02:55 PM

Search Tags


/mathhelpforum @mathhelpforum