I have a problem where I need to find where the function is increasing/decreasing, critical values, concavity etc... But I think my first derivative may be wrong.

Question: $\displaystyle g(x)=\frac{4x-2.2}{(4x-4.84)^2}$

Using the quotient formula, I have the following:

$\displaystyle g'(x)=\frac{4*(4x-4.84)^2-2(4x-4.84)*4*(4x-2.2)}{(4x-4.84)^4}$

$\displaystyle g'(x)=\frac{4*(-4x-.44)}{(4x-4.84)^3}$

So to find my critical points, I set my numerator to 0.

$\displaystyle -4x-.44=0$

$\displaystyle -.44=4x$

$\displaystyle -.11=x$

$\displaystyle 4x-4.84=0$

$\displaystyle x=1.21$

I also set my numerator to 0 (something to do with vertical asymtotes but I am not sure why we do this)

Using a number line, I think the function will be:

Decreasing:(-infinity, -.11) U (1.21, infinity)

Increasing:(-.11,1.21)

Correct?

I didn't want to go any further to test for concavity just in case my 1st derivative was wrong.

TIA