Results 1 to 7 of 7

Thread: Math homework help (should be easy for you cos I'm noob)

  1. #1
    goodieboy's Avatar
    Joined
    Apr 2012
    Posts
    2,176
    Userbars
    7
    Thanks
    364
    Thanked
    308/239
    DL/UL
    32/0
    Mentioned
    225 times
    Time Online
    42d 21h 2m
    Avg. Time Online
    14m

    Math homework help (should be easy for you cos I'm noob)

    It's the basics of Calculus, I think, need to use some differentiation

    Find the range of values of 'a' such that the gradient of the tangent to the curve y = (2x + a)/(4 - x^2) is always positive. ('a' is a constant here)

    I could get up to the point of dy/dx = (2x^2 + 2ax + 8)/((4 - x^2)^2), but can't go on after that.

    All help appreciated!
    Last edited by goodieboy; 12-10-2012 at 08:20 PM.

  2. #2
    goodieboy's Avatar
    Joined
    Apr 2012
    Posts
    2,176
    Userbars
    7
    Thanks
    364
    Thanked
    308/239
    DL/UL
    32/0
    Mentioned
    225 times
    Time Online
    42d 21h 2m
    Avg. Time Online
    14m
    Bump, still need help

  3. #3
    goodieboy's Avatar
    Joined
    Apr 2012
    Posts
    2,176
    Userbars
    7
    Thanks
    364
    Thanked
    308/239
    DL/UL
    32/0
    Mentioned
    225 times
    Time Online
    42d 21h 2m
    Avg. Time Online
    14m
    @(you need an account to see links), help me? Thanks! Will + rep.

    Still need this done

  4. #4

    Joined
    Jul 2012
    Posts
    717
    Userbars
    3
    Thanks
    105
    Thanked
    201/114
    DL/UL
    24/0
    Mentioned
    88 times
    Time Online
    22d 22h 4m
    Avg. Time Online
    7m
    The derivative you found is the slope. The question is asking about what ranges the slope is positive.




    Is it from -8 to positive infinity?

    y = (2x + a)/(4 - x)

    y prime= [(4 - x)(2) -(2x + a)(-1)] / (4 - x)^2

    y prime = 8-2x+2x+a/(4 - x)^2

    y prime = 8+a/(4 - x)^2 //denominator doesn't matter because it is squared. always positive except for a undefined at x=4

    8+a>0

    a>-8 <<---that's my conclusion
    Last edited by Graff; 12-09-2012 at 10:52 PM.

  5. #5
    Reese's Avatar
    Joined
    Dec 2011
    Posts
    2,316
    Userbars
    8
    Thanks
    1,177
    Thanked
    767/532
    DL/UL
    20/0
    Mentioned
    510 times
    Time Online
    40d 8h 14m
    Avg. Time Online
    13m
    Heh, I started writing it out but Graff beat me to it . I can go and doublecheck, but everything looks good ^_^
    I can try and explain everything if you'd like(gradient of tan=y2-21/x2-x1, there's lots of ways to look at what values are positive and what are negative, and that's usually just what makes sense to you)
    Last edited by Reese; 12-09-2012 at 11:04 PM.
    J_L_K_64 rules you.

  6. #6
    goodieboy's Avatar
    Joined
    Apr 2012
    Posts
    2,176
    Userbars
    7
    Thanks
    364
    Thanked
    308/239
    DL/UL
    32/0
    Mentioned
    225 times
    Time Online
    42d 21h 2m
    Avg. Time Online
    14m
    Dammit I made a friggin mistake when typing the question. Will + full rep to both of you if you guys can help me solve it again. I've edited the question, help me again please?


    @(you need an account to see links) @(you need an account to see links), the answer should be -4 < a < 4

  7. #7

    Joined
    Jul 2012
    Posts
    717
    Userbars
    3
    Thanks
    105
    Thanked
    201/114
    DL/UL
    24/0
    Mentioned
    88 times
    Time Online
    22d 22h 4m
    Avg. Time Online
    7m
    y = (2x + a)/(4 - x^2)

    y prime= [(4 - x^2)(2) -(2x + a)(-2x)] / (4 - x^2)^2

    y prime= 2(x^2 + xa + 4)/(4-x^2)^2

    2(x^2 + xa + 4)/(4-x^2)^2>0

    [2(x^2 + xa + 4)/(4-x^2)^2>0]*(4-x^2)^2
    [2(x^2 + xa + 4)>0]/2

    (x^2 + xa + 4)>0

    use quad formula (to find critical points)

    {-a(+/-)sqrt[a^2-(4*1*4)] }/2

    (a^2 -16)^.5 --->> [(a+4)(a-4)]^.5

    CPs @ a=-4 and 4



    yprime = 2(x^2 + xa + 4)/(4-x^2)^2

    plug in 3 values that are in between +/- infinity and the CP's to make sure that it's positive in that range only.



    Struggled a bit, sorry. This is my final answer, I'll let you do the algebra.
    Last edited by Graff; 12-10-2012 at 09:56 PM.

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •