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!

Increasing functions simple Q

  1. Jan 9, 2013 #1
    to cut it short, was asked to find the values of x for which f(x) is an increasing function
    [tex] \dfrac{(x^2 + 3)}{4x + 1} = f(x) [/tex]
    [tex] \dfrac{(4x-6)(x+2)}{(4x+1)^2} = f'(x) [/tex] so setting this to be greater than zero I get the values of x < -2, and x > 2/3 however in the answers they got x <= -2 and x >= 2/3, and they set f'(x) to be >= 0. I thought with increasing functions f'(x) is > 0?
     
  2. jcsd
  3. Jan 9, 2013 #2

    Mark44

    Staff: Mentor

    Some texts distinguish between increasing and strictly increasing. For an increasing function, if a < b, then f(a) ≤ f(b). For a function that is strictly increasing, if a < b, then f(a) < f(b).

    And similar for decreasing/strictly decreasing.
     
  4. Jan 9, 2013 #3
    oh I see, thank you!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Increasing functions simple Q
  1. Increasing functions (Replies: 1)

  2. Increasing functions (Replies: 8)

  3. Increasing function (Replies: 5)

  4. Increasing Function (Replies: 1)

  5. Simple integration Q (Replies: 1)

Loading...