Results 1 to 9 of 9
Like Tree1Thanks
  • 1 Post By hollywood

Math Help - Looking for inflection points

  1. #1
    Junior Member
    Joined
    Nov 2012
    From
    New Jersey
    Posts
    31

    Looking for inflection points

    Find the first positive x value of x for which f(x)=sin(e^x) has an inflection point. Correctly take derivatives.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Looking for inflection points

    Have you found the second derivative?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Nov 2012
    From
    New Jersey
    Posts
    31

    Re: Looking for inflection points

    Yes, it's e^x(cos(e^x)-e^xsin(e^x))
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Looking for inflection points

    Correct. Since 0<e^x for all real x, this leaves you to solve:

    f(x)=\cos(e^x)-e^x\sin(e^x)=0

    I recommend a numeric root-finding technique, such as Newton's method. Graph the function to determine a good initial guess.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Nov 2012
    From
    New Jersey
    Posts
    31

    Re: Looking for inflection points

    I'm sorry. I don't know Newton's method?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,807
    Thanks
    697

    Re: Looking for inflection points

    Hello, camjenson!

    Your second derivative is incorrect.


    Find the first positive value of x for which f(x)\:=\:\sin(e^x) has an inflection point.

    We have: . f(x) \:=\:\sin(e^x)

    Then: . f'(x) \:=\:e^x\cdot\cos(e^x)

    Hence: . f''(x) \;=\; e^x\!\cdot\!\cos(e^x) + e^x\!\cdot\![-\sin(e^x)]\!\cdot\! e^x \;=\;e^x\big[\cos(e^x)  - e^x\sin(e^x)\big]

    And we must solve: . \begin{Bmatrix}e^x \:=\:0 & [1] \\ \cos(e^x) - e^x\sin(e^x) \:=\:0 & [2] \end{Bmatrix}


    Equation [1] has no solutions.


    Equation [2] can be approximated.

    Let u = e^x
    We have: . \cos u - u\sin u \:=\:0 \quad\Rightarrow\quad u\sin u \:=\:\cos u \quad\Rightarrow\quad \frac{\sin u}{\cos u} \:=\:\frac{1}{u}
    And we can approximate a root of: . \tan u \:=\:\frac{1}{u}
    Then back-substitute: . x \,=\,\ln(u)

    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Nov 2012
    From
    New Jersey
    Posts
    31

    Re: Looking for inflection points

    WHat do you mean by back-substitute x=ln(u)? Sorry I'm a little slow...
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Looking for inflection points

    Quote Originally Posted by Soroban View Post
    ...

    Your second derivative is incorrect.
    The OP has the same result that you cite.

    Quote Originally Posted by camjenson View Post
    I'm sorry. I don't know Newton's method?
    You should be able to find it in your textbook, or online. I assume you are expected to know it, since you have been given this problem.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Mar 2010
    Posts
    993
    Thanks
    244

    Re: Looking for inflection points

    Or perhaps the professor expects you to use a graphing calculator to find the root. The calculator uses Newton's method, but you don't need to know that to plug in the functions and get your answer.

    - Hollywood
    Thanks from MarkFL
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: August 24th 2011, 11:35 AM
  2. Replies: 16
    Last Post: June 10th 2011, 06:49 AM
  3. Replies: 1
    Last Post: April 20th 2010, 07:36 PM
  4. Replies: 2
    Last Post: October 29th 2009, 08:02 PM
  5. Replies: 5
    Last Post: February 27th 2009, 07:48 PM

Search Tags


/mathhelpforum @mathhelpforum