Results 1 to 6 of 6

Math Help - 2 Questions

  1. #1
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1

    2 Questions

    These two questions are ones that I am just stuck on...

    #1. I found that

    f(x) = -5x

    but I need some help finding what

    a = ?


    Thanks for the help!

    -qbkr21
    Attached Thumbnails Attached Thumbnails 2 Questions-ff8904fe9349960e3f59f813adba1a1.png  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1

    Re:

    #2.
    Attached Thumbnails Attached Thumbnails 2 Questions-7d6d44109cd246a1a38eb6ad2a75591.png  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1

    Re:

    #2 (continued)...
    Attached Thumbnails Attached Thumbnails 2 Questions-d97d9017bd230b7f26109aaf8e556b1.png  
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Oct 2006
    Posts
    679
    Awards
    1

    Re:

    #2 (continued)...


    #2.

    a. g(-6) = ?
    b. g(-3) = ?
    c. g(0) = ?
    d. g(4) = ?
    e. The absolute maximum of g(x) occurs when x =___?___ & is equal to ___?___

    The problem I am having is graphing this sucker.

    Thanks

    -qbkr21
    Follow Math Help Forum on Facebook and Google+

  5. #5
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by qbkr21 View Post
    These two questions are ones that I am just stuck on...

    #1. I found that

    f(x) = -5x

    but I need some help finding what

    a = ?


    Thanks for the help!

    -qbkr21
    This is the second question i've seen today that reminds me of the second fundamental theorem of calculus, but it's not in the exact form it needs to be. anyway, this is what i'd do:

    since f(x) = -5x, f(t) = -5t

    \Rightarrow 1 + \int_{a}^{x} \frac {f(t)}{t^3}dt = 1 + \int_{a}^{x} \frac {-5t}{t^3}dt

    \Rightarrow 1 - 5 \int_{a}^{x} \frac {1}{t^2}dt = -5x^{-1}

    \Rightarrow \int_{a}^{x} \frac {1}{t^2}dt = \frac {1}{5} + x^{-1}

    \Rightarrow -(x)^{-1} + (a)^{-1} = \frac {1}{5} + x^{-1}

    and solve for a

    But this is speculation
    Follow Math Help Forum on Facebook and Google+

  6. #6
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by qbkr21 View Post
    #2 (continued)...
    Note that f(x) is always a constant, it's value only depends on the value of x. f(t) is no different.

    g(x) = \int_{-4}^{x} f(t)dt = [t]_{-4}^{x} = x - 4

    \Rightarrow g(-6) = 0, since x<-4

    i think you have it from here

    EDIT: o sorry, i think g(-6) = -6 - 4 = -10 i was treating g(x) as f(x)
    Last edited by Jhevon; May 24th 2007 at 12:13 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. More log questions
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 31st 2010, 04:58 PM
  2. Please someone help me with just 2 questions?
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 4th 2009, 04:55 AM
  3. Some Questions !
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 3rd 2009, 03:09 AM
  4. Replies: 4
    Last Post: July 19th 2008, 07:18 PM
  5. Replies: 3
    Last Post: August 1st 2005, 01:53 AM

Search Tags


/mathhelpforum @mathhelpforum