1. Limited time only! Sign up for a free 30min personal 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!

Help on graph

  1. Nov 6, 2007 #1
    ok i was graphing y=-e^x e=2.1783 and i got this up and down line growing from negative to positve, i have no clue what the hell im looking at, ive attached what i got tell if im right or wrong, point in the right direction if you could

    Attached Files:

  2. jcsd
  3. Nov 6, 2007 #2


    User Avatar
    Homework Helper
    Gold Member

    Take the graph of e^x and flip it upside down.
  4. Nov 6, 2007 #3
    thank you that was driving me nuts
  5. Nov 6, 2007 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    The graph you have posted is NOT [itex] - e^x [/itex]
  6. Nov 7, 2007 #5


    User Avatar
    Science Advisor

    Since you don't say HOW you got that graph, there's not much more we can tell you!
    (It looks to me something like y= excos(x).)
  7. Nov 7, 2007 #6
    It looks to me like a naive attempt at (-e)^x, judging from the oscillations. Of course, the points between the integers are all wrong. That MAY have been the logic behind it, but I don't know...
  8. Nov 7, 2007 #7
    poohbear, there is still some confusion. Are you trying to graph (-e)^x or -(e^x). There is a big difference, and writing -e^x is slightly ambiguous (although it implies the latter of the two I wrote).

    Moo Of Doom, you are right it looks like (-e)^x (with connecting lines drawn in the undefined parts) but when x is 2 f(x) should be around 7.4 and in his document 4 < f(2) < 6 which actually satisfies neither of the two possibilities mentioned.
  9. Nov 7, 2007 #8
    I thought it was odd too, but when you look at the original post, he says

    and 2.17822 is about 4.74. I think he mistyped his value for e when he plotted the points as well. Makes everything pretty confusing, doesn't it?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook