1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Limit Questioin

  1. Jan 13, 2012 #1
    The paragraph says, " Even if the function f is an everywhere differentiable function, it is still possible for f ' to be discountinuous. However, the graph of f ' can never exhibit a discountinuity of ...." picture is in paint document...

    What type of discountinuity is that? a hole discountinuity?

    second Question: Is my understanding correct? This is my explanation why f is not differentiable at some x value.

    Given
    f ' (x) =
    x , x>0
    x^2 x<0
    7 , x = 0.

    Therefore, the graph of the differnetiable function should have a similar discontinuity as the one shown in the paint document. (Determining the limit)
    lim f ' (x) as x → 0- = lim f ' (x) as x → 0+ = 0 however because lim Δx → 0 [ f(0 + Δx) - f(0) ]/Δx is equal to f ' (0) = 7 ≠ 0, therefore the function f non differentiable at x = 0 because inorder for f to be diff at 0, f'(0) must equal 0

    Is my understanding why f is not differentiable at x = 0 correct?

    Next, If I wanted to make f differentiable at x could I do this by defining one of two functions....

    f ' (x) =
    x , x>0
    x^2 x<0
    0 , x = 0.

    OR
    f ' (x) =
    x , x>0
    x^2 x<0

    Would this work to make f diff at x = 0?
     

    Attached Files:

    Last edited: Jan 13, 2012
  2. jcsd
  3. Jan 13, 2012 #2
    The answer to your last question is yes for the first function. The second function is not defined in 0.
     
    Last edited: Jan 13, 2012
  4. Jan 13, 2012 #3

    HallsofIvy

    User Avatar
    Science Advisor

    While the derivative of a function is not necessarily continuous, it must have the "intermediate value propert"- if f'(x0)= a and f'(x1)= then, between x0 and x1, f' must take on all values between a and b. From that, it follows that a derivative cannot have a "jump" discontinuity (the left and right limits exist but are different).
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...