Results 1 to 3 of 3

Math Help - Equalling Derivatives

  1. #1
    Newbie
    Joined
    May 2009
    Posts
    16

    Equalling Derivatives

    Suppose the function g is defined by
    g(x) = k sqrt(x + 1) for 0 [LaTeX ERROR: Convert failed] x [LaTeX ERROR: Convert failed] 3

    = mx + 2 for 3 < x [LaTeX ERROR: Convert failed] 5

    where k and m are constants. If g is differentiable at x = 3, what are the values of k and m.
    this is part c on a question, but i dont think you need the information from the original question.
    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Pinkk's Avatar
    Joined
    Mar 2009
    From
    Uptown Manhattan, NY, USA
    Posts
    419
    If it is given that g(x) is differentiable at x=3, then both pieces of the piecewise function must have the same values of g(3) and g'(3).

    So let's see where the two pieces are equal to each other when x=3:

    k\sqrt{3+1}=3m+2
    k=\frac{3m+2}{2}

    Making that substitution, let's see where the derivatives are equal to each other at x=3:

    \frac{\frac{3m+2}{2}}{2\sqrt{3+1}}=m
    \frac{3m+2}{8}=m
    m=\frac{2}{5}
    k=\frac{3(\frac{2}{5})+2}{2}=\frac{8}{5}

    So, g(x) is differentiable at k=\frac{8}{5},m=\frac{2}{5}. You can check your work to see that this piecewise function is continuous at x=3 with these constants and that there is a single value for g'(3) with these constants.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2009
    Posts
    16
    Quote Originally Posted by Pinkk View Post
    If it is given that g(x) is differentiable at x=3, then both pieces of the piecewise function must have the same values of g(3) and g'(3).

    So let's see where the two pieces are equal to each other when x=3:

    k\sqrt{3+1}=3m+2
    k=\frac{3m+2}{2}

    Making that substitution, let's see where the derivatives are equal to each other at x=3:

    \frac{\frac{3m+2}{2}}{2\sqrt{3+1}}=m
    \frac{3m+2}{8}=m
    m=\frac{2}{5}
    k=\frac{3(\frac{2}{5})+2}{2}=\frac{8}{5}

    So, g(x) is differentiable at k=\frac{8}{5},m=\frac{2}{5}. You can check your work to see that this piecewise function is continuous at x=3 with these constants and that there is a single value for g'(3) with these constants.
    thanks so much. its so simple once u explain it
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Derivatives and Anti-Derivatives
    Posted in the Calculus Forum
    Replies: 7
    Last Post: February 6th 2011, 06:21 AM
  2. Replies: 1
    Last Post: July 19th 2010, 04:09 PM
  3. Derivatives....again!
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 10th 2010, 07:32 AM
  4. Replies: 4
    Last Post: February 10th 2009, 09:54 PM
  5. Trig derivatives/anti-derivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 10th 2009, 01:34 PM

Search Tags


/mathhelpforum @mathhelpforum