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Local max and min (including a domain)

  1. Dec 4, 2015 #1
    I have this function: f(x) = √3x + 2Cos(x)
    I have to find the local maxima and minima,where -π ∠ or equal to x ∠ or equal to 2π.

    1: Differentiate the function given.
    f'(x) = 3/(2√x) -2Sin(x)

    2: Find where f'(x) is undefined or = 0 to get critical points.

    3: Find f''(x) [The second differential]

    4: Sub in critical values for x, if < 0 = Minima, if > 0 = Maxima.

    Question 1:What does this domain mean? and how will it affect my answer(s)?
    Question 2:Is there any easier way to find where f'(x) is undefined or 0 other than subbing in random values until you get ones that work? I always get stuck at this part despite knowing everything else, without the correct critical values it is pointless to continue with a question.
     
  2. jcsd
  3. Dec 4, 2015 #2

    PeroK

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    Is that ##\sqrt{3}x## or ##\sqrt{3x}## in your function?
     
  4. Dec 4, 2015 #3

    ##\sqrt{3x}##
     
  5. Dec 4, 2015 #4

    PeroK

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    Are you sure?
     
  6. Dec 4, 2015 #5
    A sugar I misread! its the other one.
     
  7. Dec 4, 2015 #6

    PeroK

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    Try that one. It might be a little easier.
     
  8. Dec 4, 2015 #7
    f'(x) = sqrt(3) - 2Sin(x)
     
  9. Dec 4, 2015 #8

    PeroK

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    And where is ##f'(x) = 0##?
     
  10. Dec 4, 2015 #9
    Couldn't tell you if my life depended on it.
     
  11. Dec 4, 2015 #10
    Been trying numbers such as 0, -1,-2,-3,-4 and 1,2,3,4... etc. but they are giving decimal places. Also does "undefined" just mean like it doesn't give a solution whatsoever?
     
  12. Dec 4, 2015 #11

    PeroK

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    If you re-arrange it, you get:

    ##sin(x) = \frac{\sqrt{3}}{2}##

    If that doesn't look familiar, then you could use the inverse function ##sin^{-1}##
     
  13. Dec 4, 2015 #12
    Ah yea so that is = 60 so one critical point is = 60. So there is a method to find where x = 0 thanks :) !!!
    And what about where it is undefined? :) and where does the domain come into a question like this?
     
  14. Dec 4, 2015 #13

    PeroK

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    Mathematically, the trig functions are defined in terms of radians, not degrees. Also, trig functions are periodic, so ##sin(x)## will take a given value at more than one point. This is where the domain comes in. You need to find all points where ##sin(x) = \frac{\sqrt{3}}{2}## in the given domain.

    You've got a nice, regular function for ##f'(x)## so that's not undefined anywhere. So, in this problem, there's no complexity like that to worry about.

    I'm not sure why you're interested in where ##f(x) = 0##. You don't need that to find maxima and minima.
     
  15. Dec 4, 2015 #14
    *f'(x) my bad.

    So are these steps correct?

    1: differentiate f(x)
    2: Find where f'(x) is 0 or undefined (i.e. critical value at 60 as calculated)
    3: Find f''(x)
    4: sub in critical point instead of x
    5: If greater than 0 it is a minimum, if less it is a maximum.

    But the question stated to find the "local maxima AND minima" I will only find one of these using this method right? And will the domain affect my answers?
     
  16. Dec 4, 2015 #15

    PeroK

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    That's the correct method. You will find all maxima and minima this way. The domain does affect your answer. If you look for any ##x##, your ##f(x)## here has an infinite number of maxima and minima.

    You're still working in degrees, I see!
     
  17. Dec 4, 2015 #16
    *1/3pi my bad!!! Thanks!!! :D
     
  18. Dec 5, 2015 #17

    I like Serena

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    Doesn't ##\sin x = \frac{\sqrt 3}{2}## have another solution when ##-\pi \le x \le 2\pi##?
    How about ##x=\frac{2\pi}{3}## (or ##x=120^\circ##)?
     
  19. Dec 6, 2015 #18

    HallsofIvy

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    No, it is NOT "= 60". It is equal to 60 degrees or [itex]\pi/3[/itex] or simply [itex]\pi/3[/itex]. You can drop the "radians" in such an answer but not degrees.
     
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