Results 1 to 11 of 11

Math Help - Continuity mx problem

  1. #1
    Junior Member mant1s's Avatar
    Joined
    Jun 2009
    From
    USA
    Posts
    34
    Awards
    1

    Continuity mx problem (Edited!)

    Hi Community,

    I've done this problem 3 times and am obviously forgetting something. Maybe I'm supposed to take the limit? Anyway here's the problem and my work...
    Code:
    find M so that f is continuous...
    
    piecewise:
    f(x) = mx - 11         if x< -8
             x^2 + 3x - 3  if x >= -8
    here's my work

    Code:
    f(-8-) = f(-8+)
    -8x - 11 = 8^2 - (-8(3)) - 3
    -8x - 11 = 85
    0 = -8x - 96
    96 / -8 != 12?
    As always any help is appreciated!
    Last edited by mant1s; June 11th 2009 at 12:38 PM. Reason: Wrote wrong problem by accident!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Ruun's Avatar
    Joined
    Mar 2009
    From
    North of Spain
    Posts
    129
    Thanks
    13
    Yes, you're supossed to take the limit. And for f to be continuous on -4, the must be the same (definition of continuity)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    Quote Originally Posted by mant1s View Post
    Hi Community,

    I've done this problem 3 times and am obviously forgetting something. Maybe I'm supposed to take the limit? Anyway here's the problem and my work...
    Code:
    find M so that f is continuous...
    
    piecewise:
    f(x) = mx - 9         if x< -4
             x^2 + 7x - 5  if x >= -4
    here's my work

    Code:
    f(-8-) = f(-8+)
    -8x - 11 = 8^2 - (-8(3)) - 3
    -8x - 11 = 85
    0 = -8x - 96
    96 / -8 != 12?
    As always any help is appreciated!
    why you take -8 you want to study the continuity at x=-4 not -8 so

    -4m-9=16-28-5

    4m+9=17

    4m=8\rightarrow m=2 right
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member mant1s's Avatar
    Joined
    Jun 2009
    From
    USA
    Posts
    34
    Awards
    1
    Quote Originally Posted by mant1s View Post
    Hi Community,

    I've done this problem 3 times and am obviously forgetting something. Maybe I'm supposed to take the limit? Anyway here's the problem and my work...
    Code:
    find M so that f is continuous...
    
    piecewise:
    f(x) = mx - 9         if x< -4
             x^2 + 7x - 5  if x >= -4
    here's my work

    Code:
    f(-8-) = f(-8+)
    -8x - 11 = 8^2 - (-8(3)) - 3
    -8x - 11 = 85
    0 = -8x - 96
    96 / -8 != 12?
    As always any help is appreciated!

    sorry guys I typed the problem out wrong... it is supposed to be -8 instead of -4. I accidently typed the value from a sample problem.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member mant1s's Avatar
    Joined
    Jun 2009
    From
    USA
    Posts
    34
    Awards
    1
    Quote Originally Posted by Ruun View Post
    Yes, you're supossed to take the limit. And for f to be continuous on -4, the must be the same (definition of continuity)
    Do I take the limit of each side? Sorry I'm a little confused :/
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member Ruun's Avatar
    Joined
    Mar 2009
    From
    North of Spain
    Posts
    129
    Thanks
    13
    Yes the one without m, will gave some real number k, and the second one will give you one real number depending of m. One equation, one unknowkn, solve for m, and you're ready
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member mant1s's Avatar
    Joined
    Jun 2009
    From
    USA
    Posts
    34
    Awards
    1
    I hate to be a bother but could I please have an example?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    Check this out

    We say that a function is continuos a x=c if and only if \lim_{x\to{c^-}}f(x)=\lim_{x\to{c^+}}f(x)

    So by this definition

    \lim_{x\to{-8^-}}(mx-11)=\lim_{x\to{-8^+}}(x^2+3x-3) \Longleftrightarrow{m(-8)-11}=(-8)^2+3(-8)-3

    Now solve for m
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member mant1s's Avatar
    Joined
    Jun 2009
    From
    USA
    Posts
    34
    Awards
    1
    Thanks Von,

    I solved for

    <br />
\Longleftrightarrow{m(-8)-11}=(-8)^2+3(-8)-3<br />

    by

    Code:
    -8m - 11 = -91
    -8m = -80
    m = -80/-8 ---> 10
    However the answer is wrong. Did I do my algebra right?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,922
    Thanks
    1762
    Awards
    1
    Quote Originally Posted by mant1s View Post
    Code:
    -8m - 11 = -91
    -8m = -80
    m = -80/-8 ---> 10
    However the answer is wrong. Did I do my algebra right?
    No you did not.
    Have you ever considered that may be a great deal of your trouble?
    (-8)^2+3(-8)-3=37
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member mant1s's Avatar
    Joined
    Jun 2009
    From
    USA
    Posts
    34
    Awards
    1


    Thanks Plato. I dono what my deal was on that one. I think it was just a +/- human error.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help with a continuity problem!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 26th 2011, 06:08 PM
  2. Continuity problem 2
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 2nd 2010, 08:04 PM
  3. Continuity Problem
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: March 28th 2010, 06:08 PM
  4. Continuity problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 19th 2008, 07:27 PM
  5. continuity problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: December 5th 2006, 12:34 AM

Search Tags


/mathhelpforum @mathhelpforum