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Minima, maxima of function

  1. Oct 30, 2008 #1
    How will I find minima,maxima or both of one function (if they exist)? For example:

    [tex]f(x)=x^2-3x+2[/tex];

    [tex]f(x)=x^2-2x-x+2[/tex];

    [tex]f(x)=x(x-2)-(x-2)[/tex];

    [tex]f(x)=(x-2)(x-1)[/tex]

    x1,2=1,2

    Those are the zeroes of the function.

    But how will I find the critical points (minima,maxima) of the function if [itex]x \in \mathbb{R}[/itex]
     
  2. jcsd
  3. Oct 30, 2008 #2
    The minima/maxima of a function is the point(s) where [tex]f'(x)=0[/tex]
     
  4. Oct 30, 2008 #3

    HallsofIvy

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    That's a quadratic function. Since it is concave upward it has a minimum but no maxiumum. The minimum is at the vertex you need to find the vertex. You could do that by completing the square but since you have already found the x-intercepts of the function, you can use the fact that, for a parabola with vertical axis, the vertex always lies half way between the x-intercepts.

    Since this problem has nothing to do with linear algebra or abstract algebra, I am moving it to "General Mathematics".
     
  5. Oct 30, 2008 #4
    Take the derivative of the originial function, then find where the derivative equals zero. Then plug those points (they are x-values) into your original function, and that will be your Y-coordinates. That will give you the values of your max's and min's. It will also let you see absolute max and min.
     
  6. Oct 30, 2008 #5
    Don't forget about inflection points!
     
  7. Oct 30, 2008 #6
    Do you mean like f(x)=(x-a)2+b ?

    HallsofIvy, When I draw graphic of the function it is not parabola.

    f(3)=2
    f(2)=0
    f(1)=0
    f(0)=2
    f(-1)=6
    f(-2)=12
    f(-3)=20
     
  8. Oct 30, 2008 #7
    f(x) has a relative minimum or maximum at either the endpoints of a region you're observing or when f'(x)=0. To determine whether it's a minimum, find when f''(x) is positive or negative; when f''(x) is positive, the graph is concave up, and when f''(x) is negative, the graph is concave down. Therefore, you have a relative extrema when f'(x)=0; it's a maximum if f''(x) is positive, or a minimum if f''(x) is negative.

    F''(x)=0 are your points of inflection.
     
  9. Oct 30, 2008 #8

    Eh? If inflection points are where f''(x) = 0 (which they are), what do you call points where f'(x) = 0, but is neither a maximum nor a minimum? For example, f(x) = x^3, which has no maximum or minimum has f'(0) = 0. What are those points called if it's not inflection points?
     
  10. Oct 30, 2008 #9

    HallsofIvy

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    ??? It certainly is a parabola! Any quadratic function has a parabola as its graph. And the points (3,2), (2,0), (1,0), (-1,6), (-2,12), (-3,20) definitely lie on a parabola. The vertex is, as I said before, half way between 1 and 2, at 3/2. f(3/2)= 9/4- 9/2+ 2= -9/2+ 2= -5/2. f(x)= (x- 3/2)^2- 5/2.
     
  11. Oct 30, 2008 #10

    D H

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    Saddle points.

    Edit
    Perhaps that was too hasty. The problem is that the term inflection point describes a point where the second derivative reaches this. Inflection points obviously do not in general coincide with stationary points. A saddle point is by definition a stationary point that is not a local extremum.
     
    Last edited: Oct 30, 2008
  12. Oct 30, 2008 #11
    Inflections points are the points where the 2nd derivative equals zero, at those points the concavity will change.

    Where the first derivative is zero are called critical points, those are the relative max's and mins of a function.
     
  13. Oct 30, 2008 #12

    D H

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    More appropriately, the points where the first derivative is zero are called stationary points. The absolute value function has a well-defined critical point: x=0. It has no stationary point because the derivative is non-zero everywhere except for x=0 and is not defined at x=0.
    Not all stationary points are extremal. Tac-Tics gave a perfectly good example of a function that has a stationary point but no extremal points.
     
  14. Oct 30, 2008 #13

    HallsofIvy

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    A quadratic function does not have any "inflection" points. And the original question has already been completely answered.
     
  15. Oct 30, 2008 #14
    Ah, yes. Thank you. It's been about 6 years since I studied this. The concept is there, but the terminology is fuzzy and needs dusting :-)
     
  16. Oct 30, 2008 #15
    Since all of your functions are quadratic, an alternative to calculus is to complete the square. Ie., your first function is f(x) = x^2 - 3x + 2 = (x - 3/2)^2 - 5/2. Since squares of real numbers are never negative, your function has the minimum point (3/2, -5/2).
     
  17. Oct 31, 2008 #16
    Thanks for the posts. I found the minima with derivation of the original formula. So my goal is [itex]x^2-3x+2=(x-a)^2+b[/itex]
    [tex]x^2-3x+2=x^2-2ax+a^2+b[/tex]
    [tex]-3x+2=-2ax+(a^2+b)[/tex]
    [tex]-3=-2a[/tex]
    and
    [tex]2=(a^2+b)[/tex]

    So, a= 3/2
    and
    [tex]b=2-9/4[/tex]
    [tex]b=-1/4[/tex]

    [tex] x^2-3x+2=(x-3/2)^2-1/4 \geq -1/4=f(3/2)[/tex]

    So 3/2 is the local minimum.
     
  18. Oct 31, 2008 #17

    HallsofIvy

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    In other words, the vertex of the parabola is at x= 3/2, half way between the x-intercepts. That's what I said in post 3.
     
  19. Dec 7, 2008 #18
    I am again stuck with minima and maxima of function.

    The original problem:
    [tex]f(x)=\frac{4x}{3x^2+4}[/tex]

    Now, I tried:

    [tex]f(x)=\frac{4}{3}*\frac{x}{x^2+\frac{4}{3}}=\frac{4}{3}*\frac{1}{x+\frac{4}{3x}}[/tex] and I don't know how to go on out of here. I was trying to have something in the form (x+a)2+b > or < b

    Also I tried [tex]\frac{4x}{3x^2+4} + 1 - 1=\frac{3(x^2+\frac{4}{3}x+\frac{4}{3})}{3(x^2+\frac{4}{3})} - 1 = \frac{(x-\frac{2}{3})^2+\frac{6}{9} \geq \frac{6}{9}}{(x-0)^2 + \frac{4}{3} \geq \frac{4}{3} } - 1[/tex]
    Again nothing. Please help me!

    Thanks in advance.
     
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