Results 1 to 7 of 7

Math Help - Help me with providing example

  1. #1
    Member
    Joined
    Oct 2009
    From
    Canada
    Posts
    128

    Help me with providing example

    Provide an example of function f:R --> R and g:R--> R, and c is subset to R, such that g is not differentiable at c and fog is defferentiable at c.

    Provide an example of function f:R --> R and g:R--> R, and c is subset to R,
    such that f is not differentiable at g(c) and fog is defferentiable at c.

    Provide an example of function f:R --> R and g:R--> R, and c is subset to R,
    such that f+g is differentiable at c and f is not defferentiable at c.


    it kind of makes me confused
    Follow Math Help Forum on Facebook and Google+

  2. #2
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by 450081592 View Post
    Provide an example of function f:R --> R and g:R--> R, and c is subset to R, such that g is not differentiable at c and fog is defferentiable at c.

    Provide an example of function f:R --> R and g:R--> R, and c is subset to R,
    such that f is not differentiable at g(c) and fog is defferentiable at c.

    Provide an example of function f:R --> R and g:R--> R, and c is subset to R,
    such that f+g is differentiable at c and f is not defferentiable at c.


    it kind of makes me confused
    you do mean c is an element of R, right? As in c \in \mathbb{R}.

    For the first, c = 0, f(x) = 0 and g(x) = |x|

    For the second, c = 0, f(x) = |x| and g(x) = 0

    Now try the third.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2009
    From
    Canada
    Posts
    128
    um There is a contracdiction in the second one, they are both saying |0| is differentiable and not defferentible
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Oct 2009
    From
    Canada
    Posts
    128
    actually, the second example works too, but it seems like you changing the rang of the function though, cause f and g is not an element of R anymore, it becomes {0}

    so change the constant function f(x) =0 to f(x) = x
    Last edited by 450081592; November 10th 2009 at 08:20 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by 450081592 View Post
    um There is a contracdiction in the second one, they are both saying |0| is differentiable and not differentible
    no, that is not what the example says.

    Quote Originally Posted by 450081592 View Post
    actually, the second example works too, but it seems like you changing the rang of the function though, cause f and g is not an element of R anymore, it becomes {0}

    so change the constant function f(x) =0 to f(x) = x
    we are giving two separate examples, it does not matter if the ranges differ in one example from the other.

    f and g are real functions. and 0 is an element of \mathbb{R}.

    f(x) = x does not work in the first example. neither would it work for the second example...
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Oct 2009
    From
    Canada
    Posts
    128
    Quote Originally Posted by Jhevon View Post
    no, that is not what the example says.

    we are giving two separate examples, it does not matter if the ranges differ in one example from the other.

    f and g are real functions. and 0 is an element of \mathbb{R}.

    f(x) = x does not work in the first example. neither would it work for the second example...

    oh I meant f(x) = X^2, this will work
    Follow Math Help Forum on Facebook and Google+

  7. #7
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by 450081592 View Post
    oh I meant f(x) = X^2, this will work
    yes. that will work in the first example
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Why providing a model means a theory is consistent?
    Posted in the Discrete Math Forum
    Replies: 7
    Last Post: August 29th 2011, 07:58 AM
  2. providing example of one to one ,onto functions......
    Posted in the Discrete Math Forum
    Replies: 6
    Last Post: September 24th 2010, 06:08 AM
  3. Providing Conditions
    Posted in the Differential Geometry Forum
    Replies: 10
    Last Post: February 8th 2010, 09:03 AM
  4. Replies: 0
    Last Post: June 2nd 2008, 12:17 PM

Search Tags


/mathhelpforum @mathhelpforum