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Homework Help: 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


    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!
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