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Symmetry of Cubic Function

  1. Sep 27, 2012 #1
    As we know, a quadratic function can be expressed in a form of complete square by a method of completing the square. This form enables us to prove that a quadratic equation is symmetric about its stationary point.
    But for the cubic function, is there a similar way to prove that the cubic curve is inversely symmetric about its point of inflection?? (meaning to prove that the curve on each side of the inflection point is inverted but perfectly matched )


    Thanks a lot
     
  2. jcsd
  3. Sep 27, 2012 #2

    Mentallic

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    Actually, it turns out that it does! I didn't know this before today.

    Since you mentioned the inflexion point, I'm guessing you know calculus. So the proof goes a little something like this:

    [tex]f(x)=ax^3+bx^2+cx+d[/tex]

    [tex]f'(x)=3ax^2+2bx+c[/tex]

    [tex]f''(x)=6ax+2b[/tex]

    Now obviously to find where the inflexion point is, we let the second derivative equal to 0, so we get

    [tex]6ax+2b=0[/tex]

    [tex]x=\frac{-b}{3a}[/tex]

    Now, the most important point in this proof is the part that comes next. We know that functions are odd or even based on some simple properties [itex]f(-x)=-f(x)[/itex] and [itex]f(x)=f(-x)[/itex] respectively, but this only applies to functions that are odd or even about the y-axis.

    What we're trying to prove is if all cubics are odd about their inflexion point (most of the time their inflexion point is not on the y-axis, so whenever [itex]b\neq 0[/itex]).

    Before we try to figure out what we're supposed to be evaluating, let's take a simpler example:

    [tex]g(x)=x^2-2x+1=(x-1)^2[/tex]
    How can we prove that g(x) is an even function about its turning point? Well, if we evaluated g(x+1), we will get [itex]g(x+1)=x^2[/itex] which is an even function as we know, but how do we show that?

    That is, we can't just slap the even property of f(x)=f(-x) onto g, because then we'd be trying to solve g(x+1)=g(-(x+1)) and g(-x-1) is not equal to g(x+1). So what is? Well, if it turns out that we need to solve g(x+1)=g(-x+1).
    You should try figure out why this is for yourself.

    Ok, so applying what we've discovered to the cubic, it turns out that we want to see if

    [tex]f\left(x-\frac{b}{3a}\right)=-f\left(-x-\frac{b}{3a}\right)[/tex]

    which if true, satisfies the proof that all cubics are symmetric about their inflexion point.
     
    Last edited: Sep 27, 2012
  4. Sep 28, 2012 #3
    Thank you very much. U r inspiring .
     
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