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!

Math challenge

  1. Nov 1, 2005 #1
    ok here is a good one i found
    a) f'(-1)=f'(1)=0
    b) f'(x)<0 when -1<x<1
    c) f'(x)>0 when x<-1 and x>1
    d) f(-1)=4; f(1)=0; and f(0) has no value
    e) f''(x)<0 when x<0 and f''(x)>0 when x>0
    what is the original f?
  2. jcsd
  3. Nov 2, 2005 #2


    User Avatar
    Science Advisor

    Are you sure f(-1) = 4 and not f(-1) = -4?
  4. Nov 2, 2005 #3


    User Avatar
    Science Advisor
    Homework Helper

    Are you sure your not just asking us to do your homework :uhh: ?
  5. Nov 2, 2005 #4


    User Avatar
    Science Advisor
    Homework Helper

    Does this even have a unique solution? You've mostly given inequalities, so I suspect not.

    Let's look at the interval (<,-1]
    f'(-1)=0, the graph is horizontal at x=-1
    f'(x)>0 when x<-1. It is increasing all the way to x=-1
    f(-1)=4, goes through the point (-1,4)
    e) f''(x)<0 when x<0. And is concave.

    Well, I can think of one function in this region which satisfies it. A mountain parabola like -(x-a)^2+b. The vertex should be at -1, so that gives a. It goes through (-1,4) so this gives b.
    Surely you can find more functions. It doesn't even matter whether we 'compress' the graph to the line x=-1, so -k(x-a)^2+b works just as well for any k>0.
  6. Nov 2, 2005 #5
    idk it was on a quiz in (AP calc) i had the other day and i got it wrong and was wonder if anyone could get an answer for it.
  7. Nov 2, 2005 #6


    User Avatar
    Science Advisor

    I guessed the following solution

    F(x) = c1x^-1 + c2 x + c3

    You can check it all works out, minus one problem (as I mentioned above) the condition f(-1) =4 clashes with the concavity conditions on f''.. C1 needs to be strictly positive.

    Nor would it be unique, the inverse power term works so long as its power is odd.

    so try something like

    f = x^-1 +x -2
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Math challenge
  1. Challenging integral (Replies: 25)

  2. Challenging ODE (Replies: 7)

  3. Limsupreme challenge (Replies: 1)

  4. Integral challenge (Replies: 7)

  5. Challenge Problem (Replies: 1)