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Homework Help: First-Derivative Test

  1. Nov 9, 2009 #1
    1. The problem statement, all variables and given/known data
    http://img5.imageshack.us/img5/2218/49718059.jpg [Broken]

    2. Relevant equations
    in picture

    3. The attempt at a solution
    why am i getting this wrong?
    when x=-8.8 its the abs.max value
    and x=-7.2 is the abs.min value...
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 9, 2009 #2
    -8.8 appears to be a relative min if g'(x) changes from - to +.

    Think about it down and then up: \ /

    seems like a minimum
  4. Nov 9, 2009 #3
    im a bit confused with the wording.
    relative = local or absolute?
    i assumed it is absolute.

    so i kinda get why -8.8 is min.. but then -7.2 has to be max, which would look like: /
    is that correct?
  5. Nov 9, 2009 #4
    I had a great right up for you, but when I posted it I was timed out and had to login in (drives me nuts!!!!!!) and lost everything. I don't have time to write it again, but here is a short version.

    g'(x) is

    + (-inf, -8.8)
    - (-8.8, -8)
    - (-8, -7.2) Asymptote at -8
    + (-7.2, + inf)

    I didn't observe the vertical asymptote at -8 in my earlier post. This changes the slope.

    -8.8 is actually a max

    -7.2 is a min.

    Relative implies either local or abs. Evaluating g(x) at the critical points will enable you to tell if its local or abs. Abs values will yields the very largest and very smallest values for g(x) for all values of x within the given domain. Locals give you min or max within a local range. Locals can be absolute if there is only one max and/or one min. Say you have two rel max, the point with a larger g(x) will be the abs and the smaller g(x) will be the local.
  6. Nov 10, 2009 #5

    so would that work:
    http://img29.imageshack.us/img29/5402/40917982.jpg [Broken]

    the only thing i dont get is why
    - (-8.8, -8)
    - (-8, -7.2)
    and not +?
    Last edited by a moderator: May 4, 2017
  7. Nov 10, 2009 #6


    Staff: Mentor

    g'(x) can change signs only at zeros in the numerator, which are x = -8.8 and x = -7.2. For x < -8.8, g'(x) > 0. For x > -7.2, g'(x) > 0. For -8.8 < x < -7.2, g'(x) < 0 (although g'(x) is undefined at x = -8.)
  8. Nov 11, 2009 #7
    i got it! the problem was with the + and - signs.. i had to switch places! what a stupid mistake :S
    Last edited: Nov 12, 2009
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