Results 1 to 3 of 3

Math Help - [help checking]Find a such that f'(0) exists:

  1. #1
    Member
    Joined
    May 2009
    Posts
    211

    [help checking]Find a such that f'(0) exists:

    I'm not sure if i got this right since instead of using the limit definition i just differentiated, but...

    Given the piecewise function: f(x) = sin(x) , x<= 0 ; f(x) = ax - x^2 , x>0
    find the value a such that f'(0) exists.

    What I did:
    (sinx)' = cosx and (ax-x^2)' = a-2x ; then i set them equal and solve for f'(0) to get a=1

    and that's my answer, which i think makes sense since when i graph it there doesn't seem to be any corners.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    That looks good to me.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Quote Originally Posted by Arturo_026 View Post
    I'm not sure if i got this right since instead of using the limit definition i just differentiated, but...

    Given the piecewise function: f(x) = sin(x) , x<= 0 ; f(x) = ax - x^2 , x>0
    find the value a such that f'(0) exists.

    What I did:
    (sinx)' = cosx and (ax-x^2)' = a-2x ; then i set them equal and solve for f'(0) to get a=1

    and that's my answer, which i think makes sense since when i graph it there doesn't seem to be any corners.

    Well, yes: that's the correct answer, but the reason why it is so is not so clear: the derivative f'(0) exists iff both one-sided derivatives (the left one and the right one) exist finitely and they're the same, and that's why you "can" calculate the left and the right derivatives as you did and equal them.
    But EVEN this is wrong since you're assuming by the above that the derivative not only exists but MUST be continuous at x = 0, which by no means must be so! So you've better to use the definition of the derivative f'(0) from the left (i.e., lim [f(x) - f(0)]/x, with x --> 0 from the left), then the same for the right derivative, and them equal them.
    Since your function is a nice one you''ll get the same answer as you got at the beginning (i.e., because f'(x) does happen to be continous), but this time you'll do it "by the book"

    Tonio
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: September 20th 2010, 03:48 AM
  2. Replies: 3
    Last Post: August 31st 2010, 09:19 PM
  3. Find the inverse function, if exists..
    Posted in the Calculus Forum
    Replies: 6
    Last Post: April 8th 2009, 08:43 AM
  4. Checking whether integral exists
    Posted in the Calculus Forum
    Replies: 4
    Last Post: October 25th 2008, 01:25 AM
  5. Find sum if it exists
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: December 6th 2007, 09:14 PM

Search Tags


/mathhelpforum @mathhelpforum