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!

Derivative at 0 problem

  1. Oct 28, 2013 #1
    1. The problem statement, all variables and given/known data

    f(x) = |x| + x
    Does f'(0) exist? Does f'(x) exist for values of x other than 0?
    This is from lang's a first course in calculus page 54 # 13


    2. Relevant equations
    lim (f(x+h) - f(x))/h
    h->0


    3. The attempt at a solution
    So I'm not sure if I am doing this correctly
    at first I took the derivative of f(x) = |x| + x

    f(x) = |x| + x = √x^2 + x
    from there

    f'= lim h -> 0 (√(x+h)^2 + (x+h) - (√x^2 + x))/h
    = x + h +x + h - x - x / h
    = 2h/h
    =2

    I assumed this meant that f' = 2 but this doesn't make sense because there is no slope at x = 0

    so I decided to take the right and left derivatives

    right:
    for x > 0, h>0
    |x| = x
    f'= (|x+h| + (x + h) - (|x| + x))/h = ((x+h) + (x + h) - ( x + x )) / h = 2

    left:
    for x < 0 , h < 0
    |x| = -x

    f' = (-(x+h) + (x + h) - (-x +x ))/h = -1

    so my conclusion is that there is no derivative at x = 0 because the left and right derivatives do not equal.

    I am not very confident in what I did here. If someone can help me understand it better I would really appreciate it!
     
    Last edited: Oct 28, 2013
  2. jcsd
  3. Oct 28, 2013 #2

    UltrafastPED

    User Avatar
    Science Advisor
    Gold Member

    |x| does not have a derivative at zero due to the sharp point of the "V".
     
  4. Oct 28, 2013 #3
    So how do I show that mathematically?
    I understand that geometrically, but when I do problems like these I have a tough time doing it
     
  5. Oct 28, 2013 #4
    also when I graph it out it only shows the right side / instead of V
    I don't understand how I got the left derivative as -1 when there's nothing on the left
     
  6. Oct 28, 2013 #5

    Mark44

    Staff: Mentor

    Typo - should be f(x) = |x| + x.
    You need more parentheses, as in (f(x + h) - f(x))/h
    √(x + h)2 ≠ x + h. Think about the case when x + h < 0.
    Your conclusion is correct, but there's an error in your work. If you sketch a graph of y = |x| + x for x < 0, it should be clear that the slope is 0, not -1
     
  7. Oct 28, 2013 #6
    shouldn't f'(0) = 0 or undefined or dne since there is no derivative at 0?
     
  8. Oct 28, 2013 #7

    Mark44

    Staff: Mentor

    Since the derivative doesn't exist for x = 0, the f' is not defined at 0.
     
  9. Oct 28, 2013 #8

    UltrafastPED

    User Avatar
    Science Advisor
    Gold Member

    For x > 0 |x| = x; so the slope for x>0 is 1.
    For x < 0 |x| = -x; and the slope for x<0 is -1.

    These are constants, so the limit from the left (x<0) at x=0 is -1. From the right (x>0) it is +1.
    So the left and right derivatives are not the same ... this is due to the kink at x=0.


    So for the expression g(x) = -|x| + x you get g(x) = 0 for x>= 0, and g(x) = 2x for x < 0;
    so now the slope is +2 for x<0, but it is 0 for x>0.
     
  10. Oct 28, 2013 #9

    Mark44

    Staff: Mentor

    The OP went back and fixed his first post. It should have been f(x) = |x| + x.

    For x > 0, f'(x) = 2; for x < 0, f'(x) = 0.
     
  11. Oct 28, 2013 #10
    Thanks for the replies! I fixed up the errors in my thread and added parentheses.

    so I re-did the left for x<0
    I got
    f' = (-(x+h) + (x+h) - (-x+x))/h = (-x - h + x + h + x - x)/h = 2x/h but taking the limit as h->0 it is undefined

    is it an algebraic error I am making?
     
  12. Oct 28, 2013 #11
    Yeah sorry for the confusion
     
  13. Oct 28, 2013 #12

    Mark44

    Staff: Mentor

    The numerator isn't 2x. Check your algebra.
    Yes
     
  14. Oct 28, 2013 #13
    sorry

    f' = (-(x+h) + (x+h) - (-x+x))/h = (-x - h + x + h + x - x)/h = 0/h not 2x/h
     
  15. Oct 28, 2013 #14

    Mark44

    Staff: Mentor

    Right. And in the limit, even though h → 0-, for each value of h that isn't zero, the quotient is zero, hence the limit is zero as well.

    So as you found, if x > 0, f'(x) = 2, and if x < 0, f'(x) = 0. At x = 0, f' doesn't exist.
     
  16. Oct 28, 2013 #15
    I completely understand now! thank you so much for your help and clarity, I appreciate it!
     
  17. Oct 29, 2013 #16

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    You have already done it: you showed that the left- and right-derivatives are unequal, so there cannot be a limit of [f(h) - f(0)]/h as h → 0.

    Note that h → 0 means that h is allowed to have either sign, as long as its magnitude goes to 0. If we want to specify a direction of approach to 0 we must write either h → 0+ or h → 0-.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Derivative at 0 problem
  1. Derivative = 0 (Replies: 5)

  2. Derivatives at 0 (Replies: 5)

Loading...