# Thread: DBL Check My Derivative PLS

1. ## DBL Check My Derivative PLS

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

2. algebra looks, just saw a few typos.

you mean g', not g
and you wrote 4.8 instead of 4.84 on the fifth line

3. Thanks, I've made the changes but it appears my critical numbers are wrong?

Don't I just set -4x-.44=0?

4. Ok, now I am having trouble finding my 2nd derivative and testing for concavity!

This is what I have so far:

$\displaystyle g"(x)=\frac{(4x-4.84)^3(-16)-(-4x-.44)(4)(3(4x-4.84)^2(4))}{(4x-4.84)^3}$

This is up to where I know I have not made any mistakes yet..