(adsbygoogle = window.adsbygoogle || []).push({}); 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

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Some questions about solving for critical points

**Physics Forums | Science Articles, Homework Help, Discussion**