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)
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= +/- 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
df(x) = 0 = 5[9+x^2]=45+5x^2
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...
The Attempt at a Solution