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Check if the following functions are differentiable

  1. Jan 8, 2012 #1
    1. The problem statement, all variables and given/known data
    Check if the following functions f : ℝ → ℝ are differentiable:
    [itex]\displaystyle f(x)=|(x-1)^{2}(x+1)^{3}|[/itex]

    [itex]\displaystyle f(x)=|x^{2}-\pi^{2}|sin^{2}x[/itex]

    2. Relevant equations



    3. The attempt at a solution
    I don't know what the condition should be, I've searched a lot of topics concerning this problem but they all contain reference to some points or intervals at which it should be checked.
    That's why I have no idea how to start and what definition has to be used.

    Thanks in advance!
     
  2. jcsd
  3. Jan 8, 2012 #2

    Simon Bridge

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    Do you know what a derivative is?
    What it means for the derivative to exist at a point?

    Notice that these are absolute value functions.
    What does that do to the values?
    Have you tried plotting any of them?
     
  4. Jan 8, 2012 #3
    They are "obviously" differentiable, except at a set of isolated points. Check those points. Show that the limit at x (from the definition of the derivative) does not exist. You can show this by noting that if a limit exists, it is unique.
     
  5. Jan 8, 2012 #4
    After hours of reading I finally got to that, but the problem now is how to divide [itex]\displaystyle f(x)=|(x-1)^{2}(x+1)^{3}|[/itex] into cases so that it'd be so easily calculated as eg. [itex]|x-2|[/itex] where we have 2 cases - for x > 2 and x < 2, then make the derivative of both functions and check whether it's the same or not.
     
  6. Jan 8, 2012 #5
    If you square something, it's always positive, so you can remove the absolute value sign.
     
  7. Jan 8, 2012 #6

    SammyS

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    Hello Pole . Welcome to PF !

    Have you learned about the differentiability of any broad classes of functions, such as polynomials, sinusoidal (sin(x) & cos(x)) functions, exponential functions, etc. ?

    The absolute value function, |x|, has a piecewise definition, so when it is combined with other functions, you should check at those places at which the function is 'pieced together' .
     
  8. Jan 8, 2012 #7

    SammyS

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    Almost right. Actually, if you square something, it's always non-negative (it can be zero), so you can remove the absolute value sign.

    So, the first function can be written as:
    [itex]f(x)=(x-1)^2\left|(x+1)^3\right|[/itex]​
    The piecewise definition of the absolute value function is:
    [itex]\displaystyle |x|=\left\{\matrix{x\,,\text{ if }\ x\ge0\\
    -x\,,\text{ if }\ x<0} \right.[/itex]​
     
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