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Graphing a function using critical points and increasing/decreasing intervals

  1. Jan 8, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the local maxima and minima and sketch:

    Critical points: (3, -4) and (6, 0)

    Interval of increase: (3, ∞)
    Interval of decrease (-∞, 3)

    I'm not quite sure if I graphed this correctly since I wasn't given the function to doublecheck.

    3. The attempt at a solution

    2v0fwxi.jpg


    Thanks
     
  2. jcsd
  3. Jan 8, 2013 #2

    haruspex

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    Looks ok as far as you have drawn it, but you need to decide what happens for x > 6. Please post your thoughts on that.
     
  4. Jan 8, 2013 #3
    Well since 6 is a critical point, wouldn't the curve change at that point?
     
  5. Jan 8, 2013 #4

    haruspex

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    Yes. What are the different types of critical point?
     
  6. Jan 8, 2013 #5
    I'm not quite sure, the only thing that this chapter has went over are the critical points for minimums and maximums. And I don't think I have enough information from the above to make much more determination from it?
     
  7. Jan 8, 2013 #6

    haruspex

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    From http://en.wikipedia.org/wiki/Critical_point_%28mathematics%29: [Broken]
    "a critical point of a function of a real variable is any value in the domain where either the function is not differentiable or its derivative is 0"​

    If you can't assume the function is differentiable everywhere then the question cannot be answered, so I'll assume it is. I.e. the critical points in this case are actually stationary points, so the derivative is zero. (http://en.wikipedia.org/wiki/Statio..._points.2C_critical_points_and_turning_points)
    There are four kinds of stationary point: local minimum, local maximum, rising saddle, and falling saddle. The first two are also known collectively as local extrema or turning points. At a local minimum, the gradient (as x increases) goes -ve, 0, +ve. At a local max it goes +ve, 0, -ve. At a rising saddle: +ve, 0, +ve; at a falling saddle -ve, 0, -ve.
    (A saddle is also a kind of inflection, but inflections generally include places that are not stationary points.)
    So armed with all that, what can you say about the two given points?
     
    Last edited by a moderator: May 6, 2017
  8. Jan 8, 2013 #7
    Well, since (3, ∞) is increasing, wouldn't it be fair to say that the critical point (6,0) is a rising saddle?
     
    Last edited by a moderator: May 6, 2017
  9. Jan 8, 2013 #8

    haruspex

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    Precisely.
     
  10. Jan 8, 2013 #9

    Mark44

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    (3, ∞) is just an interval, so it's incorrect to describe it as increase or decreasing. The function is increasing on (3, ∞).
     
  11. Jan 8, 2013 #10

    haruspex

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    I presumed that's what was meant, given the OP. But another point is that it should really be described as non-decreasing on that interval.
     
  12. Jan 8, 2013 #11

    Mentallic

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    Good point, but probably a minor detail for students at this level. Clarity > precision.
     
  13. Jan 8, 2013 #12
    Thank you.

    I have another question and rather than start a new thread, I think I could just post it here since its the same subject.


    I have a function: [tex]f(x) = \frac{x^2 - 1 }{x^2+1} [/tex]

    Once again the goal is to find the critical points.

    So I take the derivative of it which is:

    [tex]f'(x) = \frac{4x}{(x^2 +1)^2} [/tex]

    Now to find those points we set the left to 0 and then solve right?

    [tex] 0 = \frac{4x}{(x^2 +1)^2} [/tex]
    I'm stuck at this function and where to go how to solve since I tried to factor it but the squaring ends up canceling out any negatives.
     
  14. Jan 8, 2013 #13

    haruspex

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    Can't you just multiply out to get rid of the division?
     
  15. Jan 8, 2013 #14

    Mentallic

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    It's customary (not necessary, but common practice) to put the = 0 on the right side. So you'd have

    [tex]f'(x) = \frac{4x}{(x^2 +1)^2} [/tex]
    Setting f'(x) to 0 gives us
    [tex]\frac{4x}{(x^2 +1)^2}=0 [/tex]

    Now, if a fraction a/b equals to zero, what does this say about the fraction?
     
  16. Jan 8, 2013 #15
    Oops I forgot to include that in the above.

    [tex](x^2 + 1)^2 = 4x [/tex]

    Then square root for it to become:

    [tex] x^2+1 = 2 \sqrt[2]{x} [/tex]

    am I on the right path? I've done it on paper multiple times but I've come out with answers that don't come close to what I have of the function when it's graphed out.
     
  17. Jan 8, 2013 #16

    haruspex

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    No, that would be true if the other side of the preceding equation were 1. But it was 0.
     
  18. Jan 8, 2013 #17

    Mark44

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    Moving to Calculus & Beyond.
     
  19. Jan 8, 2013 #18
    Do you mean something like this?

    [tex] 0 = 4x (x^2 + 1)^{-2} [/tex]
     
  20. Jan 8, 2013 #19

    Mark44

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    That's not necessary. If a/b = 0, then it must be that a = 0. Can you use this idea to solve 4x/(x2 + 1)2 = 0?
     
  21. Jan 8, 2013 #20
    So [tex] 0 = \frac{4x}{(x^2 + 1)^2} [/tex]

    [tex] 0 = \frac{4(0)}{((0)^2 + 1)^2} [/tex]

    [tex] 0 = \frac{0}{(1)^2} [/tex]

    [tex] 0 = \frac{0}{1} [/tex]

    Is that what you mean?
     
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