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!

I am wondering if I am starting the curve sketching right

  1. Aug 1, 2012 #1
    1. Sketch the curve



    2. [itex]\frac{x}{(x-1)^{2}}[/itex]



    3. Here's what I have; the domain is all x except x ≠ 1, y-intercept of 0, x-intercept of 0, symmetric about the origin since it is an odd function, horizontal asymptote of 0, vertical asymptote of 1, and for the increasing and decreasing of the function, I'm wondering if I did it right. f'(x) = [itex]\frac{1*(x-1)^{2}-x*2(x-1)*1}{(x-1)^{4}}[/itex] = [itex]\frac{(x-1)^{2}-2x(x-1)}{(x-1)^{4}}[/itex] = after factoring out the one (x-1) [itex]\frac{-x-1}{(x-1)^{3}}[/itex] Increasing on (-1,1) and decreasing on (-∞,-1), (1,∞). There is no local maximum, because f(1) would result in division by 0 which is impossible. Local minimum of f(-1) = [itex]\frac{-1}{4}[/itex]
     
  2. jcsd
  3. Aug 1, 2012 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    A lot of that is right. But it's certainly not an odd function and symmetric around the origin, is it? Some of the other stuff you've said would contradict that.
     
  4. Aug 1, 2012 #3
    Based on using f(-x) would result in it being negative divided by positive which would make it an odd function, because it would be f(-x) = -f(x).
     
  5. Aug 1, 2012 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    f(1) is undefined. f(-1)=(-1/4) (as you said). That doesn't sound symmetric or odd to me.
    (x-1)^2 is not an even function.
     
  6. Aug 1, 2012 #5
    So the answer would be neither odd or even and no symmetry.
     
  7. Aug 1, 2012 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Did you sketch the graph? It doesn't look even or odd or symmetric to me. If you agree, then yes.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: I am wondering if I am starting the curve sketching right
  1. Am I doing it right? (Replies: 3)

  2. Am I right? (Replies: 3)

  3. Am i doing this right (Replies: 4)

  4. Am i right here? (Replies: 0)

  5. Am I doing this right? (Replies: 1)

Loading...