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Curve Sketching help

  1. Nov 20, 2009 #1
    Curve Sketching help!!

    1. The problem statement, all variables and given/known data
    http://img21.imageshack.us/img21/7660/54035638.jpg [Broken]


    2. Relevant equations



    3. The attempt at a solution
    For part E, ive been sitting on it for the last 2 hours trying to figure out why this is wrong..
    i came up with a solution but it doesnt seem to be right..
    if solving for f'(x) i get x=6.279
    so f'(x) > 0 ==> x<6.279
    f'(x) < 0 ==> x>6.279
    decreasing (-inf, 6.279)U(6.279, inf) and increasing on (6.279, 0) ? im not sure about the increasing part..
    but when graphing it on my calc, it look like 1.521 would work better..
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 21, 2009 #2

    Mark44

    Staff: Mentor

    Re: Curve Sketching help!!

    I drew a very rough sketch showing the two asymptotes and the zero at x = .39. For all x < .39, f(x) < 0, right? Also, f(0) [itex]\approx[/itex] -.02 from your work.

    As x --> [itex]-\infty[/itex], f(x) --> 0 and is negative. So somewhere way out to the left, the function values are less negative (larger) than the one at x = 0. This means that, going left to right, the graph of f is decreasing for a while, levels off, and then heads up to (0, -.02) and continues on up. The upshot is that the tangent line has to be horizontal at some negative value of x (not anywhere near 6.3). Check your differentation work.
     
  4. Nov 21, 2009 #3
    Re: Curve Sketching help!!

    so it would be,
    decreasing on (-inf, 0)U(0, inf) and increasing on (0, -0.02)
    right?
     
  5. Nov 21, 2009 #4

    Mark44

    Staff: Mentor

    Re: Curve Sketching help!!

    No. The graph is decreasing on (-inf, A) U (1.521, inf), and increasing on (A, 1.521). You have to find the number A, and it is less than zero.
     
  6. Nov 21, 2009 #5
    Re: Curve Sketching help!!

    the only number that is less than 0 is -0.016858 from this question.. and that doesnt work :(
     
  7. Nov 21, 2009 #6

    Mark44

    Staff: Mentor

    Re: Curve Sketching help!!

    In your OP you said that g'(6.279) = 0. That is wrong, as I said in post #2. What do you have for g'(x)?
     
  8. Nov 21, 2009 #7
    Re: Curve Sketching help!!

    for g' i get
    g'(x) = (-100x + 627) / (10x-15.21)2

    if g'(x) = 0 then x=6.27
    i tried doing the derivative 5 times already and i got the same result
     
  9. Nov 21, 2009 #8
    Re: Curve Sketching help!!

    At first glance that derivative isn't correct. Shouldn't the bottom be ^4 not ^2?
     
  10. Nov 21, 2009 #9
    Re: Curve Sketching help!!

    it wouldnt matter anyways..
    i need the top to be equal to 0.
    -100x + 627 = 0
    x=6.27
     
  11. Nov 21, 2009 #10
    Re: Curve Sketching help!!

    I don't think the top is correct either.

    How are you arriving at your derivative?
     
  12. Nov 21, 2009 #11
    Re: Curve Sketching help!!

    (f'g - g'f) / g^2

    so f(x) = 10x-3.9
    and g(x) = (10x-15.21)^2

    [10(10x-15.21)^2 ] - [2(10x-15.21)*10*(10x-3.9)] / [(10x-15.21)^2]^2
    ==> [10(10x-15.21) - 20(10x-3.9) ] / (10x-15.21)^3
    ==> (100x - 152.1 - 200x + 780) / (10x-15.21)^3
    ==> (-100x + 627) / (10x-15.21)^3
     
  13. Nov 22, 2009 #12

    Mark44

    Staff: Mentor

    Re: Curve Sketching help!!

    Error in the 3rd line: the 780 should be 78, since you have 20*3.9, not 20*39.

    So g'(x) = (-100x - 74.1)/(10x - 15.21)^3

    If you recall, I've been saying that the derivative is zero for some negative value of x.
     
  14. Nov 22, 2009 #13
    Re: Curve Sketching help!!

    oh it my mistake again.. havent paid attention to it :/
    g'(x) = 0 when x=-0.741

    so now i know that there is a local min at x=-0.741

    for part G, it will concave up (-inf, 1.521) and concave down (-0.741, 1.521)U(1.521, inf)
    the IP is x=-1.872

    right?
     
  15. Nov 22, 2009 #14

    Mark44

    Staff: Mentor

    Re: Curve Sketching help!!

    Right.
    Right.
    The inflection point seems about right - I haven't worked that out, but where you have it seems to be the right place. Your concavity descriptions are all wrong, though. The whole idea of an inflection point is that it's where the concavity changes. To the left of the inflection point, the graph is concave down. Between the inflection point and 1.521, the graph is concave up, and for x > 1.521, the graph is also concave up.
     
  16. Nov 22, 2009 #15
    Re: Curve Sketching help!!

    right right right! i got it before i read the message.. i wish i saw it earlier to save time though :b

    thank you so much once more!
     
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