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Some questions about solving for critical points

  1. Jul 18, 2011 #1
    1. The problem statement, all variables and given/known data
    I SOLVED THE MISTAKE WITH MY DERIVATIVE IN POST BELOW

    I never understood why we just get rid of the imaginary solutions?
    like for example
    if f(x) = (5*x)/(9-x^2)
    then
    df(x)/dx=(5[9+x^2])/(9-x^2)

    I've realized that when we set the numerator equal to zero we are doing this????
    df(x) = 0 = 5[9+x^2]
    what's the significance of this?
    we are setting the equation that explains the change in f(x) to be zero makes sense to find critical points

    so than what exactly is the meaning of setting the denominator equal to zero???
    like I know you can't divide by zero and what have you (for the moment) but I was wondering what would be the meaning of such a thing

    dx= 0 = 9-x^2

    setting the equation that explains the change in x to be equal to zero??? I'm trying to wrap my head around that would even mean if you could do such a thing x_x... just for the hell of it

    dx= 0 = 9-x^2
    x^2=9
    x= +/- 3
    so setting the denominator equal to zero, the equation that represents the change in x??? produces the points on the original function were f(x) is undefined... interesting... but ya

    so
    df(x) = 0 = 5[9+x^2]=45+5x^2
    -45 =x^2
    -9=x^2
    +/- 3*i=x

    f(+/- 3*i)= (+/-15*i)/(9-(+/- 3*i)^2)=(+/-15*i)/(9-(-9))=(+/-15*i)/(9+9)=(+/-15*i)/18=+/-(5*i)/6

    so these so called critical points that aren't really critical points are
    (+/- 3*i, +/-(5*i)/6)
    how would we go about plotting such points??? and what exactly do they mean... I really doubt they are meaningless...

    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Jul 18, 2011
  2. jcsd
  3. Jul 18, 2011 #2
    f'(x) ≠ 0 at any value of x means the function has no areas where it is instantaneously not changing. If this function was a function that described the position of an object, then it would mean that it's velocity is never zero. No horizontal slope of zero degrees on the function at any point in other words. I'm not sure what the imaginary numbers would get you though, not very familiar with them.

    I think you might have made a mistake when taking the derivative of the function. If you have a function f(x) = g(x)/h(x), then f'(x) = [g'(x)h(x)-h'(x)g(x)]/(h(x))^2

    If you set the denominator of f'(x) in the above equation equal to zero, the roots that you will obtain, if any, are of vertical asymptotes in the equation of f'(x). Areas where f'(x) approaches infinity or negative infinity.
     
  4. Jul 18, 2011 #3
    Indeed I did. It's suppose to be
    df(x)/dx=(5[9+x^2])/(x^2-9)

    setting the numerator equal to zero
    df(x)=0=5[9+x^2]=45+5x^2
    -45=5x^2
    -9=x^2
    +/- 3*i=x
    back into the original equation
    f(+/- 3*i)= (+/-15*i)/(9-(+/- 3*i)^2)=(+/-15*i)/(9-(-9))=(+/-15*i)/(9+9)=(+/-15*i)/18=+/-(5*i)/6
    these so called meaningless points are then
    (+/- 3*i, +/-(5*i)/6)

    back to the derivative
    df(x)/dx=(5[9+x^2])/(x^2-9)
    dx=0=x^2-9
    9=x^2
    x=+/-3
    back into the original function
    f(+/-3)=(5*x)/(9-(+/-3)^2)=(5*x)/(9-9)=(5*x)/0

    I always thought that a derivative was more of ratio, hence the notation df(x)/dx, a ratio between an infinitely small changed in f(x), to an infinitely small change in x, and we describe this ratio with a function

    If the function is a rational function of the form

    df(x)/dx = P(x)/Q(x)

    I thought that we could then say from the logic above that

    df(x) = P(x) an infinitely small change in f(x) can be described by a function P(x)
    dx=Q(x) an infinitely small change in dx can be described by a function Q(x)

    then a derivative
    df(x)/dx = P(x)/Q(x)
    and is a ratio in the form of a function that describes an infinitely small change in f(x) to x

    I assume that df(x) is not necessarily = P(x)
    nor is dx=Q(x)
    but sense it's a ratio it's sort of a simplified version of the original functions that described them but can still be said to equal to them

    and so I was just thinking if we set the denominator equal to zero of a derivative df(x)/dx
    dx = 0
    your dividing by zero... but your also set dx = 0 and finding values at which the infinitely small change in x is equal to zero which happen to be were the original function is undefined... when the infinitely small change in x values near the particular point is zero...

    Is my logic extremely flawed? Can I think of the derivative as a ratio of an infinitely small change in f(x) to x? It seems to make some sense and explains why we plug the values into the original function that are equal to when dx equals to zero is undefined... because an infinitely small change in x being equal to zero all by itself with reference to nothing else at all of just some random function really makes no sense at all... or does it X_X

    and ya I wounder what is the meaning of these "nonsense critical points"
     
    Last edited: Jul 18, 2011
  5. Jul 18, 2011 #4

    Ray Vickson

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    Science Advisor
    Homework Helper

    You say "It's suppose to be df(x)/dx=(5[9+x^2])/(x^2-9)". That is wrong; it's supposed to be df(x)/dx = 5(9+x^2)/(x^2-9)^2.

    You say
    "If the function is a rational function of the form

    df(x)/dx = P(x)/Q(x)

    I thought that we could then say from the logic above that

    df(x) = P(x) an infinitely small change in f(x) can be described by a function P(x)
    dx=Q(x) an infinitely small change in dx can be described by a function Q(x)"

    There is no logic at all here; it is just plain wrong, and makes no more sense than saying that since 1/2 = 75/150, we can think of 1 as 75 and 2 as 150.

    RGV
     
  6. Jul 18, 2011 #5
    I agree with that logic now that i think about it... I still don't like the notation that implies that there's some kind of ratio or something of the sort...

    My other wounder is when we say that function is undefined at a point we are arguing that
    the point is of the form (x, undefined)

    Could we ever encounter the opposite of such a type of a point...
    (undefined, y)

    if we wanted to evaluate a function at infinity and negative infinity we could do so... but could would we consider it of the form (undefined, y)? hmmm...

    If we were to do so we would find the derivative of the original function and divide both the denominator and numerator by the highest power in the denominator then take the limit as x-> +/- infinity

    I don't think I was ever told why we divide the numerator and the denominator by the highest power that occurs in the denominator... nor can find an answer any were when I look online... like i get that your just multiplying by 1 but why this form of it?
     
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