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Is this function differentiable at -π/2? Is it differentiable at 0?

  1. Oct 5, 2014 #1
    The function f: R → R is: f(x) =

    (tan x) / (1 + ³√x) ; for x ≥ 0,

    sin x ; for (-π/2) ≤ x < 0,

    x + (π/2) ; for x < -π/2
    _

    For -π/2 I would say it is not differentiable since both π and 2 are constants and you can not vary a constant.

    For 0, I would derivate it by using the first function - (tan x) / (1 + ³√x) - and just replace x for 0. It gives me 0 as result. If x ≠ 0, the result could also be ≠ 0.

    From here I can't go further...
     
  2. jcsd
  3. Oct 5, 2014 #2

    SteamKing

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    Homework Helper

    Your thread title is confusing. In English, we ask, "Is this function differentiable at 0?", for example.

    As for the content of your post, I believe you are being asked if the piecewise function f(x) has a derivative at certain values of x. What you are supposed to do is apply the definition of the derivative to this function at the indicated values of x and decide if the definition is satisfied. If the definition is satisfied for a certain value of x, then the function is differentiable at that point.

    The general approach is to first establish if the function is continuous at the point of interest. If it is not, no further work is needed; a function must be continuous at a point for a derivative to exist. The second part is to apply the limit definition of the derivative at the point of interest, and see if the limit exists. If the function is continuous and the limit of the derivative exists at the point of interest, then the function is differentiable at that point.

    http://en.wikipedia.org/wiki/Derivative


    This argument makes no sense. Isn't 0 a constant, like both π and 2? In any event, you are not being asked to vary these constants, merely determine if f(x) is differentiable. You can't simply pick one part of the piecewise function and decide to use that part alone to calculate the derivative or determine differentiability.

    It might be more clear if f(x) is specified as:
    Code (Text):

              x + (π/2),           for x < -π/2;
    f(x) =    sin x ,              for (-π/2) ≤ x < 0;
              (tan x) / (1 + ³√x), for x ≥ 0
     
     
  4. Oct 5, 2014 #3
    Hello SteamKing,

    Thank you for your reply! I translated it from Dutch... sorry for my mistake.
    Indeed, I am trying to check if a function f(x) has a derivative at the gives values of x.

    For -π/2, I choose f(x) = (sin (x)) and plug-in -π/2. It becomes >> f'(-π/2) = cos (-π/2) = 0,99 = 1

    For 0, I choose (tan x) / (1 + ³√x). I suppose I should derivate using the quotient rule and then plug-in 0. But then I get 0 as result.
     
  5. Oct 5, 2014 #4

    mfb

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    Where does the 0,99 come from? It is wrong.
    For this point you have to consider function values a both sides of -π/2 - the function is identical to the sine function at one side only.

    You cannot "choose" anything here. And how do you get zero?
     
  6. Oct 5, 2014 #5
    Could someone solve it for me? At least I could see it and try to understand how it has to be done.
    I really have no clue on how to solve this...
     
  7. Oct 5, 2014 #6

    Mark44

    Staff: Mentor

    No, we won't do your work for you. From the forum rules (see Homework Guidelines at https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/)
    Did you read what SteamKing and mfb wrote? They have provided some guidance for you.
     
    Last edited: Oct 5, 2014
  8. Oct 5, 2014 #7

    RUber

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    Perhaps you would find it helpful to first plot the graph. This might make the question of continuity immediately obvious. Next, you have to evaluate the derivatives of both the left side and right side functions at the point in question to see if they are the same.
    If you have trouble with the derivatives, you could always use an online tool like wolframalpha.com to check your answers.
     
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