# Compute f'(x) and finding critical numbers

• Nov 19th 2010, 07:40 AM
snypeshow
Compute f'(x) and finding critical numbers
Hey guys

Here's the solution to a question i had on a test

http://img832.imageshack.us/img832/5103/64344098.png

ok so i don't understand how they got from:
\$\displaystyle -x^3/(3x-2)(2+3x)\$
to
\$\displaystyle {3x^2(4-3x^2)}/(2-3x)^2(2+3x)^2\$

and i don't get how they find the "set numerator to 0" and "set denominator to 0" answers.

any help REALLY appreciated, thanks!
• Nov 19th 2010, 07:49 AM
harish21
This looks like a part of a test or assignment, and it is against MHF Policy to post these..

• Nov 19th 2010, 07:53 AM
Unknown008
When you differentiate this:

\$\displaystyle f(x) = \dfrac{-x^3}{(3x-2)(2+3x)}\$

you get:

\$\displaystyle f'(x) = \dfrac{3x^2(4-3x^2)}{(2-3x)^2(2+3x)^2}\$

using the quotient rule.

The denominator can be written as:

\$\displaystyle (3x-2)(2+3x) = (3x-2)(3x+2) = 9x^2 -4\$

and the quotient rule applied:
If
\$\displaystyle y = \dfrac{u}{v}\$

then;

\$\displaystyle y' = \dfrac{v u' - uv'}{v^2}\$

When you set the numerator to zero, you are looking for the maxima and minima (and/or inflexion)
When you set the denominator to zero, you are looking for the vertical asymptotes.