Results 1 to 3 of 3

Math Help - the function f(x) = e^3x +6x^2 +1 has a horizontal tangent at x =?

  1. #1
    Junior Member LexiRae's Avatar
    Joined
    Apr 2008
    Posts
    25

    the function f(x) = e^3x +6x^2 +1 has a horizontal tangent at x =?

    i cant solve it!
    im really lost. i know you find f''(x) (i got 9e^3x +12) but i dont know where to go from there
    what would i do on my calculator?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Reckoner's Avatar
    Joined
    May 2008
    From
    Baltimore, MD (USA)
    Posts
    1,024
    Thanks
    75
    Awards
    1

    Smile

    Quote Originally Posted by LexiRae View Post
    i cant solve it!
    im really lost. i know you find f''(x) (i got 9e^3x +12) but i dont know where to go from there
    what would i do on my calculator?
    A horizontal tangent will have a slope of 0. So set f'(x)=0 and solve for x.
    Last edited by Reckoner; February 25th 2009 at 04:58 PM. Reason: Stupid error. Thanks, skeeter
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,623
    Thanks
    428
    Quote Originally Posted by Reckoner View Post
    A horizontal tangent will have a slope of 0. So set f''(x)=0 and solve for x.
    set f'(x) = 0 rather than f'' ?

    f(x) = e^{3x} + 6x^2 + 1<br />

    f'(x) = 3e^{3x} + 12x = 0

    graph f'(x) and look for the zero.

    using a calculator ... x \approx -0.156
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: April 5th 2010, 04:17 AM
  2. Horizontal Tangent
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 19th 2009, 10:06 PM
  3. horizontal tangent
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 26th 2009, 08:49 PM
  4. Replies: 2
    Last Post: June 22nd 2009, 05:51 PM
  5. Replies: 5
    Last Post: January 11th 2009, 10:19 PM

Search Tags


/mathhelpforum @mathhelpforum