1. Not finding help here? Sign up for a free 30min 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!

For what values of c and d will x be differentiable for all values

  1. Oct 19, 2009 #1
    For what values of c and d will x be differentiable for all values

    [tex]
    f(x)=\left\{\begin{array}{cc}cx+d,&\mbox{ if }
    x\leq -1\\cx^3+x+2d, & \mbox{ if } x>-1\end{array}\right.
    [/tex]

    i took the derivative on each side

    f'(x) = c

    f'(x) = 3cx²+1

    c = 3cx²+1

    so i get c as -(1/2)

    but when i go get d i get two different values

    I replace X for -1 and C for -(1/2) for the first equation i get D as-(1/2) and for the second equation i get D as (1/4) what am i doing wrong?

    thanks in advance
     
    Last edited: Oct 19, 2009
  2. jcsd
  3. Oct 19, 2009 #2

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: differentiable

    Replace x for -1 and c for -(1/2 in which equations? You don't know what f(-1) is yet, so you can't solve for d simply by knowing what c is like that.

    When you took the derivative of each side, you just assumed that f(x) was continuous at x=-1 (otherwise at least one of those values would be infinite). How can you ensure this is actually true?
     
  4. Oct 19, 2009 #3
    Re: differentiable

    these equations:

    cx + d and cx³ + x + 2d

    my professor said it was continuous
     
  5. Oct 19, 2009 #4

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: differentiable

    So if I substitute in -1/2 and -1 for c and x I get

    cx+d = 1/2+d

    cx3+ x + 2d = -1/2+2d

    Nothing says that both of these need to be zero, which is what it seems you did. What condition do these have to satisfy?
     
  6. Oct 19, 2009 #5
    Re: differentiable

    i have to find the value for what c and d will x be differentiable

    from what you got above the first eq gives d = -1/2 and the second gives -1/4, doest the d value have to be equal on both equations?
     
  7. Oct 19, 2009 #6

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: differentiable

    The equations I gave don't give d equal to anything. I don't understand where you are getting this idea that the statement cx+d = 1/2+d can possibly allow me to solve for d
     
  8. Oct 19, 2009 #7
    Re: differentiable

    i was solving each equation independently, am i doing it wrong
     
  9. Oct 19, 2009 #8

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: differentiable

    What equation are you solving for d? I didn't post any equations that you can solve for d
     
  10. Oct 19, 2009 #9
    Re: differentiable

    i was solving for the top and bottom part of

    [tex]
    f(x)=\left\{\begin{array}{cc}cx+d,&\mbox{ if }
    x\leq -1\\cx^3+x+2d, & \mbox{ if } x>-1\end{array}\right.
    [/tex]


    i took the derivative of both sides

    c = 3cx²+1 giving c = -(1/2) and than i was going to replace the c value and the x value which is -1 on the original equation on the left and right side
     
    Last edited: Oct 19, 2009
  11. Oct 19, 2009 #10

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: differentiable

    What do you mean by solving for the top and bottom part of f(x)? You still haven't said exactly what equation you tried to solve, and why you tried to solve it
     
  12. Oct 19, 2009 #11
    Re: differentiable

    the question says i have to get values for c & d for x to be differentiable on all values and i was going to replace the c value on the equation

    cx+d and cx³+x+2d but seperately

    for example cx+d = -(1/2)*-1 + d which equals to d = -(1/2)

    but when i solve cx³+x+2d which gives -(1/2)(-1)³+2d <=> 1/2 + 2d <=> 2d = -(1/2) <=> d = -(1/4)
     
  13. Oct 19, 2009 #12

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: differentiable

    cx+d isn't an equation! If you know that x=-1 and c=1/2, what equation were you trying to solve?
     
  14. Oct 19, 2009 #13
    Re: differentiable

    i updated my last post, than how am i supposed to find d? thanks for helping me and i got c by taking the derivative of both sides which is

    derivative of left side = c
    derivative of right side = 3cx²+1

    c = 3cx²-1 <=> knowing that x is -1 from the piecewise function i get

    c = 3c - 1 <=> -2c = -1 <=> c = 1/2
     
  15. Oct 19, 2009 #14

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: differentiable

    Ok I see
    This just gives you d=d. It doesn't allow you to solve for d in any way.

    Recall that for a function to be differentiable, it needs to be continuous. What equation tells you that f(x) is continuous at x=-1?
     
  16. Oct 19, 2009 #15
    Re: differentiable

    i was replacing it from what it shows in the piece wise function, how shall i work this problem i am getting really confused
     
  17. Oct 19, 2009 #16

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: differentiable

    For f(x) to be differentiable at -1, it has to be continuous. What value of d makes it continuous? Hint: You need the two pieces to meet when x=-1
     
  18. Oct 20, 2009 #17
    Re: differentiable

    is the c correct?

    should i "connect" the left and right side of the piece wise function like this and resolve

    cx+d = cx³+x+2d and than substitute c and x?
     
  19. Oct 20, 2009 #18

    Office_Shredder

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Re: differentiable

    Looks good to me
     
  20. Oct 20, 2009 #19
    Re: differentiable

    so d is 1 and c is -(1/2) correct?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook