Results 1 to 8 of 8

Math Help - solving critical point problems??

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    14

    solving critical point problems??

    Find a value of "a" such that the function
    f(x) = x^2e^ax has a critical point at x = -1.



    I have no clue how to do this??
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Critical points are found by makin' f'(x)=0, do it.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2008
    Posts
    14
    i dont know how f! when theres a e????
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2008
    Posts
    14
    can any one else tell me how to slove this? Please
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2008
    Posts
    12
    First you find the derivative of it, by using the product rule. If you don't know how to do that then you should start paying attention in class.

    x^2e^{ax}

    2xe^{ax} + ae^{ax}x^2 I'm almost positive this is right, the derivative of the exponent of e may be wrong. So if it is, can someone kindly point that out.

    You can carry on from here. Clean it up a bit to make life easier, then set to equal zero and substitute x then solve for a.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    Quote Originally Posted by ab32 View Post
    Find a value of "a" such that the function
    f(x) = x^2e^ax has a critical point at x = -1.



    I have no clue how to do this??

    f(x) = x^2e^{ax}

    Use product rule

    f'(x)= 2xe^{ax} + x^2ae^{ax}

    The derivative of e is

    f(x)=e^x

    f'(x)=e^xdx
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Oct 2008
    Posts
    14
    thanks for your help i understand how to get the dervi , now how do i use that -1 critical point?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    You know you have a critical point when the derivative equals 0. The problem told you the critical point was x = -1 so plug all this into your equation

    0= 2(-1)e^{a(-1)} + (-1)^2ae^{a(-1)}

    Now just solve for a
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Critical point and Saddle Point Question
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: November 21st 2009, 08:32 AM
  2. Replies: 0
    Last Post: November 3rd 2009, 10:18 AM
  3. Critical point? Help me please
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 27th 2008, 01:00 AM
  4. what is a critical point
    Posted in the Calculus Forum
    Replies: 3
    Last Post: July 27th 2008, 09:45 PM
  5. Replies: 2
    Last Post: July 23rd 2007, 09:38 PM

Search Tags


/mathhelpforum @mathhelpforum