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Homework Help: Minimum value as x varies

  1. Mar 8, 2007 #1
    hi, I am just tackling some of the harder questions we have to do and I came across this which I have never been told how to solve and my maths teacher too busy teaching the eaiser stuff to other people to expalin this lol, annoying i know, a few of us are in this situation, anyways, hear it is, thnx.

    1. The problem statement, all variables and given/known data

    The expression x^2 - 2mx has a minimum value as x varies

    qu) Find the minimum value of x^2 - 2mx
    Give your answer in terms of m

    2. Relevant equations



    3. The attempt at a solution

    I have no idea, how to approach this. I have tried reasoning logically how it might work, but came nowhere as I dunno how the expression can have a minimum value :S is the power of 2 a clue? A few of us (in my class) have tried but no-one has the fainted idea. hope you can help

    thnx
     
  2. jcsd
  3. Mar 8, 2007 #2
    If a function have a maximum or minimum at a certain point, what can you say about the derivative at that point?




    Note that this is the PRECALCULUS fourm. Knowledge of derivatives should not be expected
    Integral
     
    Last edited by a moderator: Mar 8, 2007
  4. Mar 8, 2007 #3
    what's a derivative?
     
  5. Mar 8, 2007 #4
    A derivative is the rate of change of a quantity. A derivative is an instantaneous rate of change: it is calculated at a specific instant rather than as an average over time.
     
  6. Mar 8, 2007 #5

    Dick

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    There is a more elementary way to do problems like this. Complete the square in the expression. Can you handle this? Ie write it as (x-m)^2+SOMETHING. Can you figure out what the SOMETHING is?
     
  7. Mar 8, 2007 #6
    You can do this or find the axis of symmetry and vertex. For parabolas, you dont need to use derivatives
     
  8. Mar 8, 2007 #7
    By completing the square you're essentially doing that, it will be in vertex form. [tex]y=a(x-h)^2+k[/tex]

    Whatever makes (x-h) equal 0 is your x-coordinate, k is your y-coordinate for a min/max of a quadratic

    Or at least I believe what I said to be correct, maybe I'm wrong, dunno :-/
     
    Last edited: Mar 8, 2007
  9. Mar 8, 2007 #8
    if you change the value of "m", with it always be a "happy parabola"?

    How do you find the axis of symmetry for the following parabolas?
    y=x^2 + 4x -8
    y=x^2 + 4x

    Then, how would you find the axis of symmetry for your parabola?

    Notice that the lowest point for a "happy" parabola is always on this axis. (likewise for a "sad" parabola)
     
  10. Mar 9, 2007 #9
    the lowest point is min value let say ax^2+bx+c
    the lowest point x-coordinate is /frac{b}{-2a}
     
  11. Mar 9, 2007 #10

    Gib Z

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    Nice try at the latex dragonlorder, welcome to Pf. You forgot [ tex ] brackets though. As said before, the x value at the lowest point is [itex]\frac{-b}{2a}[/itex], to get the lowest value, they are asking for the Y value! So dont forget to substitute your x value back into your equation and get the y value.
     
  12. Mar 9, 2007 #11
    would it be (x-m)^2 + m^2?

    where would i go from there?
     
  13. Mar 9, 2007 #12

    Dick

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    I would say x^2-2mx=(x-m)^2-m^2. Now (x-m)^2 is always greater than or equal to zero, yes? So what is the minimum value?
     
  14. Mar 9, 2007 #13
    o yeh oops, lol, my bad

    would the minimum value be -m^2 then?
     
  15. Mar 9, 2007 #14

    Dick

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    You are so right.
     
  16. Mar 9, 2007 #15
    thnx for all the help buddy :D
     
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