Results 1 to 3 of 3

Math Help - calculus

  1. #1
    Junior Member
    Joined
    Dec 2008
    Posts
    37

    Angry calculus

    let f be a function defined by f(x)=
    2x- x^2 for x is less than or equal to 1
    x^2 +kx +p for x>1

    for what values of k and p will f be continuous and differentiable at x=1

    i got p=6 and k=-2 but i don't think that is correct, also shouldn't there be more than one answer? please help!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,829
    Thanks
    123
    Quote Originally Posted by andyaddition View Post
    let f be a function defined by f(x)=
    2x- x^2 for x is less than or equal to 1
    x^2 +kx +p for x>1

    for what values of k and p will f be continuous and differentiable at x=1

    i got p=6 and k=-2 but i don't think that is correct, also shouldn't there be more than one answer? please help!!
    You've got:

    f(x)=-x^2+2x~and~f'(x)=-2x+2
    and
    g(x)=x^2+kx+p

    You know f(1)=1 that means the endpoint of f is P(1,1) and the slope of the graph at P is zero.

    \lim_{x\rightarrow1}(x^2+kx+p)=1~\implies~k = -p

    \lim_{x\rightarrow1}(2x+k)=0~\implies~k = -2

    Therefore p = 2

    The equation of g: g(x) = x^2-2x+2
    Attached Thumbnails Attached Thumbnails calculus-continuous.png  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    Hi

    The continuity at x=1 is expressed by
    \lim_{x \rightarrow 1^+} f(x) = f(1)

    \lim_{x \rightarrow 1^+} (x^2 + kx + p) = 1

    1 + k + p = 1

    k + p = 0

    The differentiability at x=1 is expressed by
    \lim_{x \rightarrow 1^+} f'(x) = f'(1)

    \lim_{x \rightarrow 1^+} (2x + k) = 0

    2 + k = 0

    Then k = -2 and p = 2
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: December 13th 2011, 09:11 PM
  2. Replies: 2
    Last Post: June 25th 2010, 10:41 PM
  3. Replies: 1
    Last Post: February 11th 2010, 07:09 AM
  4. Calculus III But doesn't require Calculus :)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 19th 2009, 04:23 PM
  5. Replies: 1
    Last Post: June 23rd 2008, 09:17 AM

Search Tags


/mathhelpforum @mathhelpforum