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New to Calculus -- Increasing/Decreasing with Trig. function

  1. May 25, 2015 #1
    • Member warned about posting homework-like questions in a non-homework forum section
    Consider the function f(x) = ln (cos^2(x)) When is it increasing/decreasing?
     
    Last edited by a moderator: May 25, 2015
  2. jcsd
  3. May 25, 2015 #2

    Merlin3189

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    Not sure where calculus comes in to this? What's your problem?

    Just sketch cos(x) then cos^2(x)
    You can see the answer then, but you could continue & sketch ln(cos^2(x)) if you like.
     
  4. May 25, 2015 #3

    SteamKing

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    If the derivative f'(x) is positive, what does that tell you about f(x)? What about when f'(x) = 0? When f'(x) < 0?
     
  5. May 25, 2015 #4

    Merlin3189

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    That seems like it would be the answer for a simpler function, but I would find it easier here to just examine the function itself.
    If OP is "new to calculus", will he be able to differentiate ln()? (I had to look it up - I'm too old to calculus!)

    I had actually jumped to the assumption that OP might be differentiating a function to find turning points, then looking at the shape of the function (where increasing/decreasing) to decide what sort of turning points they were.
     
  6. May 25, 2015 #5

    SteamKing

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    It would seem this problem is geared to students who are "new to calculus", in order to illustrate some of the applications of derivatives.

    We don't know how "new" the OP is to calculus, or any other context for this problem. :sorry:

    Certainly, one can plot the function and determine by inspection where it is increasing in value or decreasing in value, if a certain range of x is specified. But if you want to show globally where the function is increasing, then the examination of the derivative can determine this without the tedious calculation of functional values. :wink:
     
  7. May 25, 2015 #6

    mathman

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    Since ln(x) is a strictly increasing function for all positive x, [itex]ln(cos^2(x))[/itex] is increasing or decreasing when [itex]cos^2(x)[/itex] is increasing or decreasing.
     
  8. May 25, 2015 #7

    Merlin3189

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    Quite.(re. Mathman)
    And with cos2(x) being periodic (and well known), you get the full answer with a simple 30sec sketch (or in-the-head sketch.).

    Edit: But I'm not saying anyone should not do it by calculus - just that it is simple by looking at the graph. I suppose these days people prefer to use intelligent calculators and get a formula, but I still visualise things.
     
    Last edited: May 25, 2015
  9. May 26, 2015 #8

    Merlin3189

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    Just thought it worth mentioning that I'm not talking about "plotting" - just sketching and no real calculation beyond thinking, "log(1) is 0, so log(smaller than 1) is negative" and "squaring numbers 0 to |1| gives positive numbers 0 to 1, mainly a bit smaller".
    Since the question is asking about qualitative features of the function, just thinking about qualitative properties of the function ("sketching") rather than calculating ("plotting") seems appropriate.
     
  10. May 26, 2015 #9

    phion

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    Here's how I do this.

    [tex]\begin{array}{|l|cr} x & 0 & \pi/4 & \pi/2 & 3\pi/4 & \pi & 5\pi/4 & 3\pi/2 & 7\pi/4 & 2\pi \\ \hline -2 & - & - & - & - & - & - & - & - & - \\ \hline sin(x) & 0 & + & + & + & 0 & - & - & - & 0 \\ \hline cos(x) & + & + & 0 & - & - & - & 0 & + & + \\ \hline f'(x) & 0 & - & 0 & + & 0 & - & 0 & + & 0\end{array}[/tex]

    As you can see with this method it is easy figure out increasing and decreasing intervals. This function increases on [itex](\pi/2, \pi)[/itex] and [itex](7\pi/4, 2\pi)[/itex], and decreases on [itex](0, \pi/2)[/itex] and [itex](\pi, 3\pi/2)[/itex].
     
  11. May 26, 2015 #10

    phion

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    Similarly, you can find when [itex]f(x)[/itex] is positive or negative, and when the function is concave up or concave down by plugging in [itex]f''(x)[/itex].
     
  12. May 26, 2015 #11

    mathman

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    [itex]cos^2(x)[/itex] oscillates between 0 and 1 with the 1's at [itex]n\pi[/itex] and the zero's half way between. That is all you need.
     
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