Results 1 to 9 of 9

Math Help - Extrema help

  1. #1
    Junior Member
    Joined
    Dec 2010
    Posts
    47

    Thumbs up Extrema help

    Hi, after a long break I've decided to look at my math hw since I go back to school tomorrow. Well, I don't understand the first problem so hopefully if I can get help with this one I can understand the rest.

    1. (noncalculator) The minimum value of f(x) = x^4 - 4x^3 + k is 7. Determine the value of k

    I know how to find min and max values, but this is different so all help is appreciated.

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Dec 2011
    Posts
    83
    Thanks
    7

    Re: Extrema help

    You know that f_{min}=7=x^4-4x^3+k for some fix value of x. What you need to determine is first the value of x when the function has its minimum value, i.e. when f'(x)=0. Plug that value of x into the equation above and you'll get k.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Extrema help

    Don't forget to test your critical values obtained from f'(x) = 0. Perhaps the easiest way is to check concavity using f''(x).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Dec 2010
    Posts
    47

    Re: Extrema help

    Quote Originally Posted by Ridley View Post
    You know that f_{min}=7=x^4-4x^3+k for some fix value of x. What you need to determine is first the value of x when the function has its minimum value, i.e. when f'(x)=0. Plug that value of x into the equation above and you'll get k.
    Hi, I'm just confused with how I'm supposed to determine the value of x when the function has its min value, for, I have a "k" in the equation which prohibits me from determining x
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Extrema help

    k is a constant...what happens to it when you differentiate?

    k merely serves to vertically translate the function, does this affect the instantaneous slope?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Dec 2010
    Posts
    47

    Re: Extrema help

    Quote Originally Posted by MarkFL2 View Post
    k is a constant...what happens to it when you differentiate?

    k merely serves to vertically translate the function, does this affect the instantaneous slope?
    Oh... completely forgot about that, so if k is a constant then the first derivative is simply 4x^3 -12x^2, and when I set that equal to zero I get x=3 and x=0

    I now test to see which value of x gives the min by plugging into the equation of second derivative which is: 12x^2 - 24x

    when i plug both of these values into the equation of second derivative i get a value of 0 for when i plug in 0 and a value of 36 when i plug in 3, since 36 is positive it indicates concave up which indicates a min, thus x=3 must be my value for x

    I then plug 3 into x for original equation and for k I get -20

    does this sound correct??
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Extrema help

    All of it is good except for finding k.

    f(3)=3^4-4\cdot3^3+k=7

    3^3\left(3-4\right)+k=7

    -3^3+k=7

    k=7+3^3=34
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Dec 2010
    Posts
    47

    Re: Extrema help

    Quote Originally Posted by MarkFL2 View Post
    All of it is good except for finding k.

    f(3)=3^4-4\cdot3^3+k=7

    3^3\left(3-4\right)+k=7

    -3^3+k=7

    k=7+3^3=34
    Oh yeah thanks for catching me on that, I merely got k - 27 = 7 and instead of doing 27+7 I did -27+7, thanks for all the help!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Extrema help

    Glad to help!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Absolute Extrema, Local Extrema
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 23rd 2009, 03:08 AM
  2. Replies: 1
    Last Post: November 18th 2009, 11:22 PM
  3. Extrema
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 26th 2008, 03:34 AM
  4. Extrema if you can
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 16th 2008, 02:26 PM
  5. rel extrema
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 5th 2007, 03:33 AM

Search Tags


/mathhelpforum @mathhelpforum