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Intervals of increasing/decreasing Using Inequalities

  1. May 15, 2015 #1
    1. The problem statement, all variables and given/known data

    Identify the open intervals on which the function ##f(x) = 12x-x^3## is increasing or decreasing

    2. Relevant equations

    ##f(x)=12x-x^3##

    ##\frac {df}{dx} = 12-3x^2 = -3(x^2 - 4)##

    3. The attempt at a solution

    I'm reading out of two textbooks. One is a Larson/Edwards text, which says to use test values upon intervals and whatnot. However, my other textbook (Serge Lang), as well as the MIT 18.01 lecture videos I am watching, don't use this method and instead utilize inequalities. I prefer this method, but am having a difficult time exercising it, as my high school did very, very little with inequalities, so my experience is minimal. Here is what I've got:

    ##f## is increasing when ##f'(x) > 0##

    ##-3(x^2 - 4) > 0##

    ##(x^2 - 4) < 0##

    ##x^2 < 4##

    ##x < \pm 2##
    ##x < 2##, ##x< -2## = ##-x > 2##

    Here is where I get stuck. I'm not sure how to make sense of this answer. I know that ##f## is increasing on the interval ##(-2, 2)## and decreasing on ##(-\infty, -2)(2, \infty)##, but don't know how to translate my answer to say this. I'm sure I'm either making a simple mistake or am just missing a bit of know-how regarding these types of problems. Any help is greatly appreciated; thanks in advance!
     
  2. jcsd
  3. May 15, 2015 #2
    You've made a mistake, think again about what you wrote after x^2 < 4
     
  4. May 15, 2015 #3

    statdad

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    When you go from [itex] x^2 < 4 [/itex] to [itex] x < \pm 2 [/itex] you introduce a problem: notice that [itex] -5 < -2 [/itex] is true but [itex] (-5)^2 < 4 [/itex] is not.

    Start with
    [tex]
    x^2 - 4 < 0 \Rightarrow (x-2)(x+2) < 0
    [/tex]

    and look at the different cases to solve the inequality.
     
  5. May 15, 2015 #4
    I had noticed this, and immediately thought about absolute values, which is when I realized I was truly lost. I appreciate the insight!!
     
  6. May 15, 2015 #5

    Ray Vickson

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    Think about the graph ##y = x^2##. For what x-region do you have ##y < 4##?
     
  7. May 15, 2015 #6

    SammyS

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    However, you can work with ##\displaystyle\ x^2 < 4 \ ## more directly.

    Take the square root of both sides. (Yes, that preserves the inequality.)

    ##\displaystyle\ \sqrt{x^2\ } < \sqrt{4\ } \ ##​

    Strictly speaking, ##\displaystyle\ \sqrt{x^2\ } = \left|x\right| \ ##, so you have the following inequality.

    ##\displaystyle\ \left|x\right|<2 \ ##​
     
  8. May 15, 2015 #7
    I ended up doing this:

    ##x^2 < 4##

    ## |x| < 2 ##, giving me ##-2 < x < 2##

    I'm currently working on:​

    ##f(x) = \cos\frac {x}{2}, 0 < x < 2\pi##​
    So far, I have:

    ##\frac {df}{dx} = -\frac{1}{2} \sin\frac {x}{2}##

    ##-\frac{1}{2} \sin\frac {x}{2} > 0##

    ## \sin\frac {x}{2} > 0##​

    ## \arcsin\frac {x}{2} > \arcsin 0##​

    ##\frac {x}{2} > 0##

    ##x > 0##

    Again, I'm confused. I know that, on the interval, f(x) is never increasing.
     
  9. May 15, 2015 #8

    SammyS

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    Start a new thread for a new problem, unless what you're adding is just a very closely related extension of the original problem.

    In this case, best to start a new thread , and state the problem completely in the body of the opening post - no matter how descriptive the title of the thread.
     
  10. May 15, 2015 #9

    Ray Vickson

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    Last edited: May 15, 2015
  11. May 15, 2015 #10

    Drakkith

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    As SammyS said, please start a new thread for new problems. It keeps thread discussions from getting mixed up between different problems.
     
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