1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculus 1 help needed quickly

  1. Apr 24, 2013 #1
    1. The problem statement, all variables and given/known data


    calc3_zps3dd48d27.png
    calc2_zpsc786a589.png
    calc1_zps15908ccd.png





    3. The attempt at a solution
    I figured out #11, could use help with the other questions. I have a final tomorrow in the morning and my professor gave us a few questions to try out so help would be appreciated. For #8 I'm not sure how to draw the second derivative graph, for #12 I don't know where to begin, and for #15 I don't know how to find the anti derivative of f'(x) so I can't start the problem. Help is appreciated
     
  2. jcsd
  3. Apr 24, 2013 #2
    For 8 to draw f'(x) you need to see where the slopes are increasing, decreasing or remaining constant. From 0 to 3, you can see the slope is constant. From 3 to 6-7ish the slopes are increasing and from 7 on the slope is decreasing. You can draw a graph based on the knowledge I just told you. Do the same to that graph to get the 2nd derivative graph.
     
  4. Apr 24, 2013 #3
    That's what I pretty much thought, thanks for the reasurement. The only one I really don't got a clue on is #12, so if someone could explain that for me that'd be great.
     
  5. Apr 24, 2013 #4

    Mentallic

    User Avatar
    Homework Helper

    For 8) once you draw the derivative f '(x) then just draw the derivative of that graph to find f ''(x).


    For 12) remember that you can split an integral up into sections, for example,
    [tex]\int _ 1 ^5 f(x)dx = \int _ 1 ^2 f(x)dx + \int _2 ^ 4 f(x)dx + \int _4 ^5 f(x)dx[/tex]
    so you should do the same, choosing your sections based on the changing slope of the graph of f.


    For 15) you aren't expected to take the anti-derivative. This should be obvious because it's giving you a lot more information than just the derivative function.
    So to start drawing the function, begin with points that you can quickly determine, such as f(0) = 1. Once you draw this point on your axis, also plot the limits at the ends of your graph.

    Now, what can we determine from the derivative and second derivative? Well, f ''(x) has a perfect square in the denominator so it's always positive (except at [itex]x=\pm 2[/itex]) and so since the numerator is -8x, this means that for positive x values, the second derivative is negative, and for negative values it's positive. This tells you what the concavity of the graph is (think of the shape of a parabola x2 that has positive concavity versus a parabola -x2 which has negative concavity).

    What about the first derivative? We already know there's a problem at [itex]x=\pm 2[/itex] because the denominator is 0 at those points, so what is

    [tex]\lim_{x\to 2^-} f'(x)[/tex]

    and

    [tex]\lim_{x\to 2^+} f'(x)[/tex]

    and the same for -2?
     
  6. Apr 24, 2013 #5
    calc5_zpsb15d2686.png
    would the answer to this just be 0.5(4+6+4)=7?
     
  7. Apr 24, 2013 #6
    So for #12, the answer for g(1) would be -1 because the slope is -1, the anti derivative of that is -x and just plug in 1 and you get the answer? Also how would you find the answer for something like g(3) where the point is between the changing slopes.

    I'm kind of lost on #15 now, will give a try though. Thanks for the help.
     
  8. Apr 25, 2013 #7

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    No. g(1) ≠ -1 .

    If ##\displaystyle \ g(x)=\int_{1}^{x}f(x)\,dx\,,\ ## then ##\displaystyle \ g(1)=\int_{1}^{1}f(x)\,dx\ .\ ##

    Think of the definite integral as the area under the curve -- or the negative of the area above the curve.
     
  9. Apr 25, 2013 #8
    ok i think i got it. so i dont really even need to take the antiderivative..just look at the graph and calculate area. g(1) = 0
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Calculus 1 help needed quickly
  1. Need quick help (Replies: 2)

  2. Quick calculus help (Replies: 3)

  3. Calculus 1 help? (Replies: 3)

  4. Calculus 1 Help! (Replies: 3)

  5. Calculus 1 help (Replies: 23)

Loading...