Results 1 to 3 of 3

Thread: Equalling Derivatives

  1. #1
    Newbie
    Joined
    May 2009
    Posts
    16

    Equalling Derivatives

    Suppose the function g is defined by
    g(x) = k$\displaystyle sqrt(x + 1)$ for 0 $\displaystyle \le $ x $\displaystyle \le $ 3

    = mx + 2 for 3 < x $\displaystyle \le $ 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 $\displaystyle g(x)$ is differentiable at $\displaystyle x=3$, then both pieces of the piecewise function must have the same values of $\displaystyle g(3)$ and $\displaystyle g'(3)$.

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

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

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

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

    So, $\displaystyle g(x)$ is differentiable at $\displaystyle k=\frac{8}{5},m=\frac{2}{5}$. You can check your work to see that this piecewise function is continuous at $\displaystyle x=3$ with these constants and that there is a single value for $\displaystyle g'(3)$ 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 $\displaystyle g(x)$ is differentiable at $\displaystyle x=3$, then both pieces of the piecewise function must have the same values of $\displaystyle g(3)$ and $\displaystyle g'(3)$.

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

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

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

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

    So, $\displaystyle g(x)$ is differentiable at $\displaystyle k=\frac{8}{5},m=\frac{2}{5}$. You can check your work to see that this piecewise function is continuous at $\displaystyle x=3$ with these constants and that there is a single value for $\displaystyle g'(3)$ 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: Feb 6th 2011, 06:21 AM
  2. Replies: 1
    Last Post: Jul 19th 2010, 04:09 PM
  3. Derivatives....again!
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Apr 10th 2010, 07:32 AM
  4. Replies: 4
    Last Post: Feb 10th 2009, 09:54 PM
  5. Trig derivatives/anti-derivatives
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Feb 10th 2009, 01:34 PM

Search Tags


/mathhelpforum @mathhelpforum